Chapter 16 Playing with Numbers (Additional Questions)

Solution:

The given number is $78$.

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The general form of a two-digit number is expressing it as the sum of the products of its digits with their respective place values.

For a two-digit number, if the digit in the tens place is $a$ and the digit in the units place is $b$, the number is $10 \times a + 1 \times b$.

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In the number $78$, the digit in the tens place is $7$ and the digit in the units place is $8$.

Applying the general form formula, the number $78$ can be written as:

$10 \times 7 + 1 \times 8$

This simplifies to:

$70 + 8$

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Now, let's examine the given options:

(A) $7 \times 1 + 8 \times 10 = 7 + 80 = 87$. This is not the number $78$.

(B) $7 \times 10 + 8 \times 1$. This is exactly the general form we derived. It equals $70 + 8 = 78$.

(C) $70 + 8$. This is the expanded form, which is equivalent to the general form $10 \times 7 + 1 \times 8$. It equals $78$.

(D) Both (B) and (C).

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Since both option (B) and option (C) are correct ways to represent the number $78$ in its general or expanded form, the option that includes both of them is the most appropriate answer.

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Thus, the correct answer is (D) Both (B) and (C).

Solution:

The given 3-digit number is represented by its digits $x$, $y$, and $z$ at the hundreds, tens, and units place, respectively.

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The general form of a number expresses it as the sum of the products of each digit with its corresponding place value.

For a 3-digit number with digits $x$ at the hundreds place, $y$ at the tens place, and $z$ at the units place, the place values are:

Hundreds place: $100$

Tens place: $10$

Units place: $1$

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To write the number $xyz$ in the general form, we multiply each digit by its place value and sum the results:

$x \times \text{(Place value of hundreds)} + y \times \text{(Place value of tens)} + z \times \text{(Place value of units)}$

$x \times 100 + y \times 10 + z \times 1$

This can be written as:

$100x + 10y + z$

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Comparing this form with the given options:

(A) $x + y + z$ is the sum of the digits, not the number.

(B) $10x + 100y + z$ incorrectly assigns place values to $x$ and $y$.

(C) $100x + 10y + z$ matches the derived general form.

(D) $x \times y \times z$ is the product of the digits, not the number.

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Therefore, the correct general form of the 3-digit number $xyz$ is $100x + 10y + z$.

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Thus, the correct answer is (C) $100x + 10y + z$.

Solution:

Let the original 2-digit number be represented by $ab$. Here, $a$ is the digit in the tens place and $b$ is the digit in the units place.

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In the general form, the value of the original number $ab$ is:

$10 \times a + 1 \times b = 10a + b$

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When the digits are reversed, the new number is $ba$. Here, $b$ is the digit in the tens place and $a$ is the digit in the units place.

In the general form, the value of the reversed number $ba$ is:

$10 \times b + 1 \times a = 10b + a$

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The difference between the original number and the reversed number is calculated as:

Difference = (Original Number) - (Reversed Number)

Difference = $(10a + b) - (10b + a)$

Difference = $10a + b - 10b - a$

Group like terms:

Difference = $(10a - a) + (b - 10b)$

Difference = $9a - 9b$

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We can factor out $9$ from the expression:

Difference = $9(a - b)$

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Since $a$ and $b$ are integers (digits), their difference $(a-b)$ is also an integer. The expression $9(a-b)$ shows that the difference between the original number and the reversed number is always a multiple of $9$.

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A number that is a multiple of $9$ is always divisible by $9$.

Let's consider an example. If the number is 52, $a=5$ and $b=2$. The original number is $52$. The reversed number is $25$.

Difference = $52 - 25 = 27$.

Using the formula $9(a-b)$: $9(5-2) = 9(3) = 27$.

The number $27$ is divisible by $9$.

Consider another example. If the number is 81, $a=8$ and $b=1$. The original number is $81$. The reversed number is $18$.

Difference = $81 - 18 = 63$.

Using the formula $9(a-b)$: $9(8-1) = 9(7) = 63$.

The number $63$ is divisible by $9$.

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Based on the general form and examples, the difference $9(a-b)$ is always divisible by $9$.

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Comparing this with the given options, the difference is always divisible by:

(A) 9 - Correct, as the difference is a multiple of 9.

(B) 11 - Not always true (e.g., 21 - 12 = 9, which is not divisible by 11).

(C) 10 - Not always true (e.g., 31 - 13 = 18, which is not divisible by 10).

(D) 1 - True, any integer is divisible by 1, but 9 is a more specific factor based on the problem structure.

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Thus, the difference between the original number and the reversed number is always divisible by 9.

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The correct answer is (A) 9.

Solution:

Let the original 3-digit number be $abc$, where $a$, $b$, and $c$ are the digits in the hundreds, tens, and units place, respectively. Note that for a 3-digit number, $a$ must be non-zero ($a \neq 0$).

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In the general form, the value of the original number $abc$ is:

$100 \times a + 10 \times b + 1 \times c = 100a + 10b + c$

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When the digits are reversed, the new number is $cba$. Here, $c$ is the digit in the hundreds place, $b$ is the digit in the tens place, and $a$ is the digit in the units place. For $cba$ to be a 3-digit number, $c$ must be non-zero ($c \neq 0$).

In the general form, the value of the reversed number $cba$ is:

$100 \times c + 10 \times b + 1 \times a = 100c + 10b + a$

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The difference between the original number and the reversed number is calculated as:

Difference = (Original Number) - (Reversed Number)

Difference = $(100a + 10b + c) - (100c + 10b + a)$

Difference = $100a + 10b + c - 100c - 10b - a$

Group like terms:

Difference = $(100a - a) + (10b - 10b) + (c - 100c)$

Difference = $99a + 0b - 99c$

Difference = $99a - 99c$

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We can factor out $99$ from the expression:

Difference = $99(a - c)$

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Since $a$ and $c$ are digits (integers), their difference $(a-c)$ is also an integer. The expression $99(a-c)$ shows that the difference between the original number and the reversed number is always a multiple of $99$.

A number that is a multiple of $99$ is always divisible by $99$.

Let's consider an example. If the number is 452, $a=4, b=5, c=2$. The original number is $452$. The reversed number is $254$.

Difference = $452 - 254 = 198$.

Using the formula $99(a-c)$: $99(4-2) = 99(2) = 198$.

The number $198$ is divisible by $99$ ($198 \div 99 = 2$).

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Since $99 = 9 \times 11$, any number divisible by $99$ is also divisible by $9$ and $11$. However, the expression directly shows the difference is a multiple of $99$. Among the given options, $99$ is the largest and most direct divisor derived from the formula.

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Comparing this with the given options:

(A) 9 - Yes, $99(a-c)$ is divisible by 9.

(B) 11 - Yes, $99(a-c)$ is divisible by 11.

(C) 99 - Yes, $99(a-c)$ is divisible by 99.

(D) 101 - No.

While the difference is also divisible by 9 and 11, it is always divisible by 99, which is a more specific and complete answer among the options.

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Thus, the difference between the original number and the reversed number is always divisible by 99.

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The correct answer is (C) 99.

Solution:

The given puzzle is an addition problem involving two 2-digit numbers, $AB$ and $BA$, where $A$ and $B$ represent single digits.

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We can represent the numbers $AB$ and $BA$ in their general forms:

The number $AB$ has digit $A$ in the tens place and digit $B$ in the units place. Its value is $10 \times A + 1 \times B = 10A + B$.

The number $BA$ has digit $B$ in the tens place and digit $A$ in the units place. Its value is $10 \times B + 1 \times A = 10B + A$.

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The addition puzzle can be written as an equation:

$(10A + B) + (10B + A) = 132$

Combine the terms involving $A$ and the terms involving $B$:

$(10A + A) + (B + 10B) = 132$

$11A + 11B = 132$

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Factor out the common factor $11$ from the left side of the equation:

$11(A + B) = 132$

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To find the value of $(A + B)$, divide both sides of the equation by $11$:

$A + B = \frac{132}{11}$

Performing the division:

$132 \div 11 = 12$

So, we have:

$A + B = 12$

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Alternatively, we can solve this using column addition logic.

In the units column, we add $B$ and $A$. The unit digit of the sum is 2. This means $A + B$ is either 2 or 12 (since $A$ and $B$ are digits, $0 \leq A, B \leq 9$, so $0 \leq A+B \leq 18$). If $A+B=2$, there is no carry-over to the tens column. If $A+B=12$, there is a carry-over of 1 to the tens column.

In the tens column, we add $A$ and $B$, plus any carry-over from the units column. The sum is represented by '13' (digit 3 in the tens place and carry-over 1 to the hundreds place).

Case 1: $A + B = 2$. Carry-over from units is 0. Tens column: $A + B + 0 = 2 + 0 = 2$. This should result in 13 (or 3 with carry 1). $2 \neq 13$. So this case is incorrect.

Case 2: $A + B = 12$. Carry-over from units is 1. Tens column: $A + B + 1 = 12 + 1 = 13$. The unit digit is 3 (matching the tens digit in 132), and there is a carry-over of 1 to the hundreds place. This matches the hundreds digit in 132.

Thus, $A + B$ must be $12$.

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The value of $A + B$ is $12$.

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Comparing with the given options:

(A) 11

(B) 12

(C) 13

(D) 10

The value $12$ matches option (B).

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The correct answer is (B) 12.

Solution:

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The question asks about the divisibility rule for the number 3.

The divisibility rule for 3 states that a positive integer is divisible by 3 if and only if the sum of its digits is divisible by 3.

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The rule works both ways:

1. If a number is divisible by 3, then the sum of its digits is divisible by 3.

2. If the sum of the digits of a number is divisible by 3, then the number is divisible by 3.

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The question provides the first part of the rule: "If a number is divisible by 3...". According to the rule, the consequence is that "the sum of its digits is divisible by 3".

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Let's consider an example.

Consider the number $51$. It is divisible by $3$ because $51 \div 3 = 17$. The sum of its digits is $5 + 1 = 6$. The sum of the digits, $6$, is divisible by $3$ ($6 \div 3 = 2$).

Consider the number $123$. It is divisible by $3$ because $123 \div 3 = 41$. The sum of its digits is $1 + 2 + 3 = 6$. The sum of the digits, $6$, is divisible by $3$ ($6 \div 3 = 2$).

Consider the number $72$. It is divisible by $3$ because $72 \div 3 = 24$. The sum of its digits is $7 + 2 = 9$. The sum of the digits, $9$, is divisible by $3$ ($9 \div 3 = 3$). Note that $9$ is also divisible by $9$, but the rule guarantees divisibility by $3$.

Consider the number $10$. It is not divisible by $3$. The sum of its digits is $1 + 0 = 1$. The sum of the digits, $1$, is not divisible by $3$.

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Based on the divisibility rule of 3, if a number is divisible by 3, then the sum of its digits is also divisible by 3.

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Comparing this conclusion with the given options:

(A) 3 - This matches the divisibility rule.

(B) 6 - Not necessarily. For example, the sum of digits of 72 is 9, which is divisible by 3 but not by 6.

(C) 9 - Not necessarily. For example, the sum of digits of 51 is 6, which is divisible by 3 but not by 9.

(D) 12 - Not necessarily. For example, the sum of digits of 51 is 6, which is divisible by 3 but not by 12.

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Therefore, if a number is divisible by 3, the sum of its digits is always divisible by 3.

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The correct answer is (A) 3.

Solution:

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The question asks for the condition under which a number is divisible by $9$.

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The divisibility rule for $9$ states that a positive integer is divisible by $9$ if and only if the sum of its digits is divisible by $9$.

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This rule is a standard test for divisibility by $9$.

Let's check the given options based on this rule:

(A) The last digit is $9$. This is incorrect. For example, the number $19$ has its last digit as $9$, but $19$ is not divisible by $9$. The number $18$ is divisible by $9$, but its last digit is $8$.

(B) The sum of the digits is divisible by $9$. This is correct. This matches the divisibility rule for $9$. For example, for the number $54$, the sum of digits is $5 + 4 = 9$. Since $9$ is divisible by $9$, the number $54$ is divisible by $9$ ($54 \div 9 = 6$). For the number $162$, the sum of digits is $1 + 6 + 2 = 9$. Since $9$ is divisible by $9$, the number $162$ is divisible by $9$ ($162 \div 9 = 18$). For the number $729$, the sum of digits is $7 + 2 + 9 = 18$. Since $18$ is divisible by $9$, the number $729$ is divisible by $9$ ($729 \div 9 = 81$).

(C) The number is even. This is incorrect. For example, the number $9$ is divisible by $9$ but it is an odd number. The number $27$ is divisible by $9$ but it is an odd number.

(D) The last two digits form a number divisible by $9$. This is incorrect. This is similar to the rule for divisibility by $4$ (last two digits are divisible by 4) or $25$ (last two digits are $00, 25, 50, 75$), but not for $9$. For example, the number $109$. The last two digits form the number $09$, which is divisible by $9$. However, $109$ is not divisible by $9$.

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Therefore, a number is divisible by $9$ if and only if the sum of its digits is divisible by $9$.

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The correct answer is (B) The sum of the digits is divisible by $9$.

Solution:

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The question asks for the condition under which a number is divisible by $6$.

A number is divisible by a composite number if and only if it is divisible by all the prime factors of that composite number, considering their powers.

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The prime factorization of $6$ is $2 \times 3$.

Therefore, a number is divisible by $6$ if and only if it is divisible by both of its prime factors, $2$ and $3$.

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Let's check the given options:

(A) 2 and 3: This matches the requirement derived from the prime factorization of 6. A number divisible by both 2 and 3 is divisible by 6.

(B) 2 and 4: If a number is divisible by 4, it is already divisible by 2. So, this condition is equivalent to being divisible by 4. Numbers divisible by 4 are not necessarily divisible by 6 (e.g., 4, 8, 16). For a number to be divisible by 6, it must be divisible by 3 as well.

(C) 3 and 4: If a number is divisible by both 3 and 4, it is divisible by their least common multiple, which is $12$. Any number divisible by $12$ is also divisible by $6$. However, this is a stronger condition than necessary. For example, $18$ is divisible by $6$ and $3$, but not by $4$. The rule for divisibility by 6 specifically uses the prime factors 2 and 3.

(D) 2 and 5: If a number is divisible by both 2 and 5, it is divisible by their least common multiple, which is $10$. Numbers divisible by 10 are not necessarily divisible by 6 (e.g., 10, 20, 40). For a number to be divisible by 6, it must be divisible by 3 as well.

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Based on the divisibility rule derived from prime factorization, a number is divisible by 6 if and only if it is divisible by both 2 and 3.

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The correct answer is (A) 2 and 3.

Solution:

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We need to find the remainder when the number $5234$ is divided by $5$.

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The divisibility rule for $5$ states that a number is divisible by $5$ if and only if its last digit is $0$ or $5$.

To find the remainder when a number is divided by $5$, we only need to consider the last digit of the number.

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The given number is $5234$.

The last digit of $5234$ is $4$.

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The remainder when $5234$ is divided by $5$ is the same as the remainder when its last digit, $4$, is divided by $5$.

When we divide $4$ by $5$, we get:

$4 = 5 \times 0 + 4$

The quotient is $0$ and the remainder is $4$.

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Thus, the remainder when $5234$ is divided by $5$ is $4$.

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Comparing this remainder with the given options:

(A) 0

(B) 1

(C) 2

(D) 4

The remainder $4$ matches option (D).

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The correct answer is (D) 4.

Solution:

The given puzzle is a multiplication problem: $2 A \times A = 1 5 6$. Here, $2A$ represents a two-digit number where the tens digit is $2$ and the units digit is $A$. $A$ is a single digit.

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We can write the two-digit number $2A$ in its general form:

$2A = 2 \times 10 + A \times 1 = 20 + A$

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The single digit $A$ can be represented simply as $A$.

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Now, we can write the given multiplication puzzle as an algebraic equation:

$(20 + A) \times A = 156$

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Expand the left side of the equation:

$20A + A^2 = 156$

Rearrange the terms to form a quadratic equation:

$A^2 + 20A - 156 = 0$

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We need to find an integer value for $A$ between $0$ and $9$ (since $A$ is a digit) that satisfies this equation. We can solve this quadratic equation or test the options.

Let's try factoring the quadratic equation. We look for two numbers that multiply to $-156$ and add up to $20$. The factors of $156$ include $(6, 26)$. The difference between $26$ and $6$ is $20$. To get a sum of $+20$ and a product of $-156$, the numbers are $+26$ and $-6$.

So, we can factor the equation as:

$(A + 26)(A - 6) = 0$

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This gives two possible solutions for $A$:

$A + 26 = 0 \implies A = -26$

$A - 6 = 0 \implies A = 6$

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Since $A$ must be a single digit (from $0$ to $9$) and also greater than $0$ for the number $2A$ to be a two-digit number starting with $2$ when multiplied by $A$ to get $156$, the value $A = 6$ is the only valid solution from the options provided.

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Let's verify the answer $A=6$ by substituting it back into the original puzzle:

If $A = 6$, the number $2A$ is $26$. The puzzle becomes $26 \times 6$.

Calculate the product:

$26 \times 6 = (20 + 6) \times 6 = 20 \times 6 + 6 \times 6 = 120 + 36 = 156$.

This matches the result $156$ given in the puzzle.

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Alternatively, we can test the given options:

If $A=4$, $24 \times 4 = 96$. Not 156.

If $A=6$, $26 \times 6 = 156$. Correct.

If $A=8$, $28 \times 8 = 224$. Not 156.

If $A=2$, $22 \times 2 = 44$. Not 156.

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Both the algebraic solution and testing the options confirm that the value of $A$ is $6$.

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Comparing with the given options, $A=6$ corresponds to option (B).

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The correct answer is (B) 6.

Solution:

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The given number is $31Z5$, which is a four-digit number where $Z$ is a digit in the hundreds place.

The problem states that the number $31Z5$ is a multiple of $9$.

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According to the divisibility rule for $9$, a number is divisible by $9$ if and only if the sum of its digits is divisible by $9$.

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Let's find the sum of the digits of the number $31Z5$. The digits are $3$, $1$, $Z$, and $5$.

Sum of digits = $3 + 1 + Z + 5$

Sum of digits = $9 + Z$

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For the number $31Z5$ to be a multiple of $9$, the sum of its digits $(9 + Z)$ must be divisible by $9$.

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The variable $Z$ represents a single digit, which means $Z$ can take any integer value from $0$ to $9$ ($0 \leq Z \leq 9$).

We need to find the value(s) of $Z$ in this range such that $9 + Z$ is a multiple of $9$.

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The multiples of $9$ are $0, 9, 18, 27, ...$

Let's check possible values for $9 + Z$ based on the range of $Z$:

Since $0 \leq Z \leq 9$, the sum $9 + Z$ will be in the range:

$9 + 0 \leq 9 + Z \leq 9 + 9$

$9 \leq 9 + Z \leq 18$

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We need to find the multiples of $9$ that fall within the range $[9, 18]$. The multiples of $9$ in this range are $9$ and $18$.

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Case 1: $9 + Z = 9$

$Z = 9 - 9$

$Z = 0$

Since $0$ is a digit between $0$ and $9$, $Z=0$ is a possible value.

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Case 2: $9 + Z = 18$

$Z = 18 - 9$

$Z = 9$

Since $9$ is a digit between $0$ and $9$, $Z=9$ is a possible value.

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Any other multiple of $9$ ($0, 27, ...$) would result in a value of $Z$ outside the range $[0, 9]$.

For example, if $9 + Z = 0$, then $Z = -9$ (not a digit).

If $9 + Z = 27$, then $Z = 18$ (not a digit).

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Thus, the possible values for $Z$ are $0$ and $9$.

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Comparing the possible values of $Z$ with the given options:

(A) 0 - Possible.

(B) 9 - Possible.

(C) Both 0 and 9 - This option includes both possible values.

(D) Only 9 - Incorrect, as 0 is also possible.

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Therefore, the possible values of $Z$ are $0$ and $9$.

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The correct answer is (C) Both 0 and 9.

Solution:

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The original 2-digit number is given in its general form as $10a + b$. Here, $a$ is the digit in the tens place and $b$ is the digit in the units place.

The number obtained by reversing the digits is $10b + a$. Here, $b$ is the digit in the tens place and $a$ is the digit in the units place.

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We need to find the sum of these two numbers:

Sum = (Original Number) + (Reversed Number)

Sum = $(10a + b) + (10b + a)$

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Combine the terms involving $a$ and the terms involving $b$:

Sum = $(10a + a) + (b + 10b)$

Sum = $11a + 11b$

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We can factor out the common factor $11$ from the expression:

Sum = $11(a + b)$

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The sum of the two numbers is $11(a+b)$. This expression shows that the sum is always a multiple of $11$.

Therefore, the sum of the two numbers is always divisible by $11$.

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Let's consider an example. Let the number be 23 ($a=2, b=3$).

Original number = $10(2) + 3 = 20 + 3 = 23$.

Reversed number = $10(3) + 2 = 30 + 2 = 32$.

Sum = $23 + 32 = 55$.

Using the formula $11(a+b)$: $11(2+3) = 11(5) = 55$.

The number $55$ is divisible by $11$ ($55 \div 11 = 5$).

Consider another example. Let the number be 71 ($a=7, b=1$).

Original number = $10(7) + 1 = 70 + 1 = 71$.

Reversed number = $10(1) + 7 = 10 + 7 = 17$.

Sum = $71 + 17 = 88$.

Using the formula $11(a+b)$: $11(7+1) = 11(8) = 88$.

The number $88$ is divisible by $11$ ($88 \div 11 = 8$).

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Based on the general form and examples, the sum $11(a+b)$ is always divisible by $11$.

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Comparing this with the given options:

(A) 9 - Not always true (e.g., sum is 55, not divisible by 9).

(B) 10 - Not always true (e.g., sum is 55, not divisible by 10).

(C) 11 - Correct, as the sum is always a multiple of 11.

(D) $a+b$ - The sum is $11(a+b)$, which means the sum is a multiple of $a+b$. However, the question asks what it is *always* divisible by among the options. While it is divisible by $a+b$, it is also always divisible by 11, which is one of the choices. Option (C) is a fixed number based on the structure, whereas $a+b$ varies depending on the original digits. The question asks for a common divisor.

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Therefore, the sum of the original number and the reversed number is always divisible by 11.

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The correct answer is (C) 11.

Solution:

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We need to evaluate the truthfulness of the Assertion (A) and the Reason (R) and determine if R correctly explains A.

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Assertion (A): The number 135 is divisible by 3 and 5.

To check divisibility by 3, we sum the digits of 135:

$1 + 3 + 5 = 9$

Since the sum of the digits, 9, is divisible by 3 ($9 \div 3 = 3$), the number 135 is divisible by 3.

To check divisibility by 5, we look at the last digit of 135. The last digit is 5.

Since the last digit is 5, the number 135 is divisible by 5.

Both conditions for divisibility by 3 and 5 are met.

Therefore, Assertion (A) is True.

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Reason (R): A number is divisible by 3 if the sum of its digits is divisible by 3, and by 5 if its last digit is 0 or 5.

The first part of the reason states the divisibility rule for 3: "A number is divisible by 3 if the sum of its digits is divisible by 3". This is the standard and correct divisibility rule for 3.

The second part of the reason states the divisibility rule for 5: "and by 5 if its last digit is 0 or 5". This is the standard and correct divisibility rule for 5.

Therefore, Reason (R) is True.

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Now, let's determine if Reason (R) is the correct explanation for Assertion (A).

Assertion (A) claims that 135 is divisible by 3 and 5. Our verification of Assertion (A) involved applying the divisibility rules for 3 and 5 (sum of digits for 3, and last digit for 5).

Reason (R) explicitly states these exact divisibility rules. Therefore, Reason (R) provides the correct criteria or rules that explain why Assertion (A) is true.

Thus, R is the correct explanation of A.

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Based on the evaluation:

Assertion (A) is True.

Reason (R) is True.

Reason (R) is the correct explanation of Assertion (A).

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Comparing this with the given options:

(A) Both A and R are true, and R is the correct explanation of A. This matches our findings.

(B) Both A and R are true, but R is not the correct explanation of A.

(C) A is true, but R is false.

(D) A is false, but R is true.

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The correct answer is (A) Both A and R are true, and R is the correct explanation of A.

Solution:

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We need to match each divisibility property with its corresponding rule.

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(i) Divisible by 2: A number is divisible by 2 if its last digit is an even number, i.e., 0, 2, 4, 6, or 8. This matches rule (d).

So, (i) $\to$ (d).

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(ii) Divisible by 5: A number is divisible by 5 if its last digit is either 0 or 5. This matches rule (c).

So, (ii) $\to$ (c).

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(iii) Divisible by 10: A number is divisible by 10 if its last digit is 0. This matches rule (b).

So, (iii) $\to$ (b).

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(iv) Divisible by 3: A number is divisible by 3 if the sum of its digits is divisible by 3. This matches rule (a).

So, (iv) $\to$ (a).

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Putting the matches together, we get the following pairs:

(i)-(d)

(ii)-(c)

(iii)-(b)

(iv)-(a)

---

Now, let's compare these pairs with the given options:

(A) (i)-(d), (ii)-(c), (iii)-(b), (iv)-(a) - Matches our derived pairs.

(B) (i)-(c), (ii)-(d), (iii)-(b), (iv)-(a) - Incorrect.

(C) (i)-(d), (ii)-(b), (iii)-(c), (iv)-(a) - Incorrect.

(D) (i)-(a), (ii)-(b), (iii)-(c), (iv)-(d) - Incorrect.

---

Therefore, the correct matching corresponds to option (A).

---

The correct answer is (A) (i)-(d), (ii)-(c), (iii)-(b), (iv)-(a).

Solution:

The given addition puzzle is:

$A 7$

+ $8 B$

-----

$1 5 0$

---

Here, $A$ and $B$ are digits, meaning they are integers from $0$ to $9$.

We analyze the addition column by column, starting from the units column.

---

In the units column, we are adding the digits $7$ and $B$. The unit digit of the sum is $0$.

This means that the sum $7 + B$ must end in $0$. Possible values for $7 + B$ are $10, 20, 30, ...$

Since $B$ is a single digit ($0 \leq B \leq 9$), the sum $7 + B$ must be within the range:

$7 + 0 \leq 7 + B \leq 7 + 9$

$7 \leq 7 + B \leq 16$

---

Within the range $[7, 16]$, the only number that ends in $0$ is $10$.

Therefore, the sum of the digits in the units column must be $10$.

$7 + B = 10$

---

Solving for $B$:

$B = 10 - 7$

$B = 3$

---

Since $B=3$ is a single digit ($0 \leq 3 \leq 9$), this is a valid value for $B$.

When $7 + 3 = 10$, the unit digit is $0$ (which matches the unit digit of $150$), and there is a carry-over of $1$ to the tens column.

---

Now, let's look at the tens column to find the value of $A$ (although the question only asks for $B$).

In the tens column, we add $A$, $8$, and the carry-over $1$ from the units column. The result is $5$, with a carry-over of $1$ to the hundreds column (to make the final sum $150$).

$A + 8 + 1 = 15$

$A + 9 = 15$

$A = 15 - 9$

$A = 6$

Since $A=6$ is a single digit ($0 \leq 6 \leq 9$), this is a valid value for $A$. Note that $A$ cannot be $0$ if $A7$ is a two-digit number, and $6$ is indeed non-zero.

---

Let's verify the full addition with $A=6$ and $B=3$:

$67 + 83$

Units column: $7 + 3 = 10$ (write down $0$, carry over $1$).

Tens column: $6 + 8 + 1$ (carry-over) $= 15$ (write down $5$, carry over $1$).

Hundreds column: $1$ (carry-over) $= 1$ (write down $1$).

The sum is $150$, which matches the given puzzle.

---

The question asks for the value of $B$. We found $B=3$.

---

Comparing our result with the given options:

(A) 0

(B) 3

(C) 10

(D) 13 (but B must be a single digit)

The value $B=3$ matches option (B).

---

The correct answer is (B) 3.

Solution:

This question refers to the addition puzzle from Question 15:

$A 7$

+ $8 B$

-----

$1 5 0$

---

From the units column in the previous question, we determined that $7 + B$ must result in a number ending in $0$, which leads to $7 + B = 10$. This gives $B=3$, and there is a carry-over of $1$ to the tens column.

---

Now we analyze the tens column addition. We add the digits in the tens place ($A$ and $8$) and the carry-over ($1$) from the units column. The result's tens digit is $5$, and the sum $150$ indicates a carry-over of $1$ to the hundreds place.

So, the sum of the digits in the tens column plus the carry-over from the units column is $15$ (since $5$ is the digit in the tens place of the result, and $1$ is carried over to the hundreds place).

The addition in the tens column is:

$A + 8 + (\text{carry-over from units}) = \text{result in tens column}$

From the units column, the carry-over is $1$. The result in the tens column is $15$ (meaning 5 in the tens place and 1 carried to the hundreds place).

So, we have the equation:

$A + 8 + 1 = 15$

---

Simplify and solve for $A$:

$A + 9 = 15$

$A = 15 - 9$

$A = 6$

---

Since $A$ is a digit ($0 \leq A \leq 9$), $A=6$ is a valid digit. Also, for $A7$ to be a 2-digit number, $A$ must be non-zero, which $6$ is.

---

Let's verify the full addition with $A=6$ and $B=3$:

$67 + 83$

Units: $7+3 = 10$. Write $0$, carry $1$.

Tens: $6+8+1 = 15$. Write $5$, carry $1$.

Hundreds: $1$ (carry) = $1$. Write $1$.

The sum is $150$, which matches the puzzle.

---

The value of $A$ is $6$.

---

Comparing our result with the given options:

(A) 5

(B) 6

(C) 7

(D) 8

The value $A=6$ matches option (B).

---

The correct answer is (B) 6.

Solution:

---

This question refers to the addition puzzle from Question 15 and Question 16:

$A 7 + 8 B = 1 5 0$

where $A$ and $B$ are digits.

---

From our solutions to Question 15 and Question 16, we found the values of the digits $A$ and $B$ that satisfy the puzzle.

In Question 15, we determined that $B = 3$.

In Question 16, we determined that $A = 6$.

---

The two numbers in the puzzle are given as $A7$ and $8B$. These are two-digit numbers where $A$ is the tens digit of the first number and $B$ is the units digit of the second number.

---

Substituting the values of $A=6$ and $B=3$ into the number representations:

The first number $A7$ becomes $67$.

The second number $8B$ becomes $83$.

---

Let's verify if the sum of these two numbers is indeed $150$:

$67 + 83 = 150$

This matches the result given in the puzzle.

---

Therefore, the two numbers in the puzzle are $67$ and $83$.

---

Comparing our result with the given options:

(A) 67 and 83 - Matches our determined numbers.

(B) 57 and 83 - The first number's tens digit is incorrect ($A=5$ instead of $6$).

(C) 67 and 84 - The second number's units digit is incorrect ($B=4$ instead of $3$).

(D) 57 and 84 - Both numbers are incorrect.

---

The correct pair of numbers is $67$ and $83$.

---

The correct answer is (A) 67 and 83.

Solution:

---

The question asks for the number by which a number is divisible if it is divisible by both $2$ and $5$.

---

A property of divisibility states that if a number is divisible by two coprime numbers, then it is also divisible by their product.

Two numbers are coprime (or relatively prime) if their greatest common divisor (GCD) is $1$.

---

Let's consider the numbers $2$ and $5$. Both $2$ and $5$ are prime numbers.

The factors of $2$ are $1$ and $2$.

The factors of $5$ are $1$ and $5$.

The only common factor is $1$. Thus, the greatest common divisor of $2$ and $5$ is $1$.

Therefore, $2$ and $5$ are coprime numbers.

---

Since the number is divisible by two coprime numbers ($2$ and $5$), it must be divisible by their product.

The product of $2$ and $5$ is $2 \times 5 = 10$.

So, if a number is divisible by both $2$ and $5$, it is divisible by $10$.

---

Let's verify this with an example.

Consider the number $20$. $20$ is divisible by $2$ ($20 \div 2 = 10$) and $20$ is divisible by $5$ ($20 \div 5 = 4$). $20$ is also divisible by $10$ ($20 \div 10 = 2$).

Consider the number $30$. $30$ is divisible by $2$ ($30 \div 2 = 15$) and $30$ is divisible by $5$ ($30 \div 5 = 6$). $30$ is also divisible by $10$ ($30 \div 10 = 3$).

---

Now, let's check the given options:

(A) 7 - Not necessarily (e.g., 20 is divisible by 2 and 5, but not by 7).

(B) 10 - Correct, based on the property of coprime divisors and their product.

(C) 15 - Not necessarily. For a number to be divisible by 15, it must be divisible by 3 and 5 (since $15 = 3 \times 5$). Divisibility by 2 does not imply divisibility by 3. For example, 20 is divisible by 2 and 5, but not by 3 or 15.

(D) 25 - Not necessarily. For a number to be divisible by 25, its last two digits must be 00, 25, 50, or 75. Divisibility by 2 and 5 only guarantees the last digit is 0, which makes it divisible by 10, but not necessarily by 25. For example, 10 is divisible by 2 and 5, but not by 25.

---

Therefore, if a number is divisible by both 2 and 5, it is divisible by 10.

---

The correct answer is (B) 10.

Solution:

---

We need to determine which of the given numbers is divisible by $4$.

---

The divisibility rule for $4$ states that a number is divisible by $4$ if and only if the number formed by its last two digits is divisible by $4$.

---

Let's apply this rule to each option:

(A) 1357: The last two digits form the number $57$. Let's check if $57$ is divisible by $4$.

$57 \div 4 = 14$ with a remainder of $1$. ($4 \times 14 = 56$).

Since $57$ is not divisible by $4$, the number $1357$ is not divisible by $4$.

---

(B) 2468: The last two digits form the number $68$. Let's check if $68$ is divisible by $4$.

$68 \div 4 = 17$. ($4 \times 17 = 68$).

Since $68$ is divisible by $4$, the number $2468$ is divisible by $4$.

---

(C) 3082: The last two digits form the number $82$. Let's check if $82$ is divisible by $4$.

$82 \div 4 = 20$ with a remainder of $2$. ($4 \times 20 = 80$).

Since $82$ is not divisible by $4$, the number $3082$ is not divisible by $4$.

---

(D) 4190: The last two digits form the number $90$. Let's check if $90$ is divisible by $4$.

$90 \div 4 = 22$ with a remainder of $2$. ($4 \times 22 = 88$).

Since $90$ is not divisible by $4$, the number $4190$ is not divisible by $4$.

---

Based on the divisibility rule for $4$, only the number $2468$ is divisible by $4$.

---

The correct answer is (B) 2468.

Solution:

---

The sentence to complete is: "A number whose last two digits form a number divisible by 4 is divisible by _________."

---

This statement describes a well-known divisibility rule for a specific number.

The divisibility rule for 4 states that a number is divisible by 4 if and only if the number formed by its last two digits is divisible by 4.

---

Comparing the given sentence with the standard divisibility rule for 4, we see that the sentence is the rule itself.

Therefore, if the number formed by the last two digits of a number is divisible by 4, the number itself is divisible by 4.

---

Let's consider an example. The number 1516. The last two digits form the number 16. Since 16 is divisible by 4 ($16 \div 4 = 4$), the number 1516 should be divisible by 4. $1516 \div 4 = 379$.

Consider another example. The number 2308. The last two digits form the number 08. Since 08 (which is 8) is divisible by 4 ($8 \div 4 = 2$), the number 2308 should be divisible by 4. $2308 \div 4 = 577$.

---

Thus, the sentence should be completed with the number 4.

The complete sentence is: "A number whose last two digits form a number divisible by 4 is divisible by 4."

---

Comparing this with the given options:

(A) 2 - If a number is divisible by 4, it is also divisible by 2. But 4 is the more specific rule stated.

(B) 4 - This matches the standard divisibility rule.

(C) 8 - Not necessarily. For example, 12 is divisible by 4, but not by 8.

(D) 12 - Not necessarily. For example, 4 is divisible by 4, but not by 12. 20 is divisible by 4, but not by 12.

---

The correct answer is 4.

---

The correct answer is (B) 4.

Solution:

---

The given number is $18X6$, where $X$ is a digit (an integer from $0$ to $9$).

The problem states that the number $18X6$ is a multiple of $4$, which means it is divisible by $4$.

---

We use the divisibility rule for $4$. This rule states that a number is divisible by $4$ if and only if the number formed by its last two digits is divisible by $4$.

---

In the number $18X6$, the last two digits are $X$ and $6$. The number formed by these two digits is $X6$.

The value of the number $X6$ can be written in the general form as $10 \times X + 6$.

---

For the number $18X6$ to be divisible by $4$, the number $X6$ (or $10X + 6$) must be divisible by $4$.

We need to find the value(s) of $X$ from the given options such that $10X + 6$ is divisible by $4$.

---

Let's test each option for $X$:

(A) $X = 0$: The number formed by the last two digits is $06$. The value is $10(0) + 6 = 6$. Is $6$ divisible by $4$? $6 \div 4 = 1$ with a remainder of $2$. So, $6$ is not divisible by $4$. Thus, $X=0$ is not a possible value.

---

(B) $X = 1$: The number formed by the last two digits is $16$. The value is $10(1) + 6 = 10 + 6 = 16$. Is $16$ divisible by $4$? $16 \div 4 = 4$ with a remainder of $0$. So, $16$ is divisible by $4$. Thus, $X=1$ is a possible value.

---

(C) $X = 3$: The number formed by the last two digits is $36$. The value is $10(3) + 6 = 30 + 6 = 36$. Is $36$ divisible by $4$? $36 \div 4 = 9$ with a remainder of $0$. So, $36$ is divisible by $4$. Thus, $X=3$ is a possible value.

---

(D) $X = 5$: The number formed by the last two digits is $56$. The value is $10(5) + 6 = 50 + 6 = 56$. Is $56$ divisible by $4$? $56 \div 4 = 14$ with a remainder of $0$. So, $56$ is divisible by $4$. Thus, $X=5$ is a possible value.

---

Based on our analysis, possible values for $X$ from the options are $1, 3,$ and $5$. The question asks for "a possible value", and option (B) provides $1$, which is a valid possible value. Options (C) and (D) also provide valid possible values, but in a multiple-choice question format where only one option is typically selected, we choose one of the valid ones. Option (B) is the first listed valid possibility.

---

The correct answer is (B) 1.

Solution:

---

We need to evaluate the truthfulness of Assertion (A) and Reason (R) and determine if R correctly explains A.

---

Assertion (A): If a number is divisible by 9, it is also divisible by 3.

Let $N$ be a number that is divisible by 9.

This means $N$ can be written as $N = 9k$ for some integer $k$.

Since $9 = 3 \times 3$, we can write $N$ as:

$N = (3 \times 3) \times k = 3 \times (3k)$

Let $m = 3k$. Since $k$ is an integer, $m$ is also an integer.

So, $N = 3m$. This shows that $N$ is a multiple of 3.

Thus, any number divisible by 9 is also divisible by 3.

Assertion (A) is True.

---

Reason (R): 9 is a multiple of 3.

A number is a multiple of 3 if it can be obtained by multiplying 3 by an integer.

We can write $9 = 3 \times 3$. Here, the integer is 3.

Therefore, 9 is a multiple of 3.

Reason (R) is True.

---

Now, let's determine if Reason (R) is the correct explanation for Assertion (A).

Assertion (A) states that divisibility by 9 implies divisibility by 3. Reason (R) provides the relationship between 9 and 3, stating that 9 is a multiple of 3 (or equivalently, 3 is a factor of 9).

As shown in the explanation for Assertion (A), the fact that $9$ can be factored as $3 \times 3$ is exactly why a number divisible by $9$ ($N = 9k$) can be rewritten as a multiple of $3$ ($N = 3 \times (3k)$).

So, Reason (R) directly explains the relationship that leads to the conclusion in Assertion (A).

Therefore, R is the correct explanation of A.

---

Based on our analysis:

Assertion (A) is True.

Reason (R) is True.

Reason (R) is the correct explanation of Assertion (A).

---

Comparing this with the given options:

(A) Both A and R are true, and R is the correct explanation of A. This matches our conclusion.

(B) Both A and R are true, but R is not the correct explanation of A.

(C) A is true, but R is false.

(D) A is false, but R is true.

---

The correct answer is (A) Both A and R are true, and R is the correct explanation of A.

Solution:

---

We need to identify the numbers from the given options that are divisible by both $3$ and $5$.

---

A number is divisible by $3$ if the sum of its digits is divisible by $3$.

A number is divisible by $5$ if its last digit is $0$ or $5$.

---

For a number to be divisible by both $3$ and $5$, it must satisfy both these conditions.

---

Let's check each option:

(A) 45:

Divisibility by 3: Sum of digits = $4 + 5 = 9$. Since $9$ is divisible by $3$ ($9 \div 3 = 3$), $45$ is divisible by $3$.

Divisibility by 5: The last digit is $5$. Since the last digit is $5$, $45$ is divisible by $5$.

Since $45$ is divisible by both $3$ and $5$, it is one of the correct options.

---

(B) 120:

Divisibility by 3: Sum of digits = $1 + 2 + 0 = 3$. Since $3$ is divisible by $3$ ($3 \div 3 = 1$), $120$ is divisible by $3$.

Divisibility by 5: The last digit is $0$. Since the last digit is $0$, $120$ is divisible by $5$.

Since $120$ is divisible by both $3$ and $5$, it is one of the correct options.

---

(C) 235:

Divisibility by 3: Sum of digits = $2 + 3 + 5 = 10$. Since $10$ is not divisible by $3$ ($10 \div 3 = 3$ remainder $1$), $235$ is not divisible by $3$.

Divisibility by 5: The last digit is $5$. Since the last digit is $5$, $235$ is divisible by $5$.

Since $235$ is not divisible by $3$, it is not divisible by both $3$ and $5$.

---

(D) 300:

Divisibility by 3: Sum of digits = $3 + 0 + 0 = 3$. Since $3$ is divisible by $3$ ($3 \div 3 = 1$), $300$ is divisible by $3$.

Divisibility by 5: The last digit is $0$. Since the last digit is $0$, $300$ is divisible by $5$.

Since $300$ is divisible by both $3$ and $5$, it is one of the correct options.

---

(E) 555:

Divisibility by 3: Sum of digits = $5 + 5 + 5 = 15$. Since $15$ is divisible by $3$ ($15 \div 3 = 5$), $555$ is divisible by $3$.

Divisibility by 5: The last digit is $5$. Since the last digit is $5$, $555$ is divisible by $5$.

Since $555$ is divisible by both $3$ and $5$, it is one of the correct options.

---

The numbers that are divisible by both 3 and 5 are 45, 120, 300, and 555.

---

Therefore, the correct options are (A), (B), (D), and (E).

Solution:

---

The account balance is a 3-digit number of the form $PQR$. This means the value of the number is $100P + 10Q + R$, where $P, Q,$ and $R$ are digits, and $P \neq 0$.

The given condition is that the last digit, $R$, is either $0$ or $5$.

---

We need to determine which of the given numbers (2, 3, 5, or 9) the balance ($PQR$) is divisible by, based on the condition that $R$ is $0$ or $5$.

---

Let's recall the divisibility rules for the options:

Divisibility by 2: A number is divisible by 2 if its last digit is even (0, 2, 4, 6, or 8).

Divisibility by 3: A number is divisible by 3 if the sum of its digits is divisible by 3.

Divisibility by 5: A number is divisible by 5 if its last digit is 0 or 5.

Divisibility by 9: A number is divisible by 9 if the sum of its digits is divisible by 9.

---

The given condition is precisely the divisibility rule for 5.

If the last digit $R$ is $0$ or $5$, then by the rule for divisibility by 5, the number $PQR$ is divisible by 5.

---

Let's check if the condition implies divisibility by other numbers in the options:

Divisibility by 2: If $R=0$, the number is divisible by 2. But if $R=5$, the number is not divisible by 2 (since 5 is odd). The condition $R \in \{0, 5\}$ does not guarantee divisibility by 2.

Divisibility by 3: The sum of digits is $P+Q+R$. If $R=0$, the sum is $P+Q$. If $R=5$, the sum is $P+Q+5$. Neither of these sums is guaranteed to be divisible by 3 just because $R$ is 0 or 5. For example, if the number is 105 ($P=1, Q=0, R=5$), the sum is $1+0+5=6$, which is divisible by 3. If the number is 115 ($P=1, Q=1, R=5$), the sum is $1+1+5=7$, which is not divisible by 3.

Divisibility by 9: The sum of digits is $P+Q+R$. Similar to divisibility by 3, the sum is not guaranteed to be divisible by 9 just because $R$ is 0 or 5.

---

The condition that the last digit $R$ is $0$ or $5$ directly corresponds to the divisibility rule for 5.

---

Therefore, if the last digit $R$ is 0 or 5, the balance $PQR$ is divisible by 5.

---

Comparing with the given options:

(A) 2

(B) 3

(C) 5

(D) 9

The number 5 matches our conclusion.

---

The correct answer is (C) 5.

Solution:

---

The account balance is a 3-digit number $PQR$, where $P, Q, R$ are its digits.

The value of the number is $100P + 10Q + R$.

The problem states that the sum of the digits, $P+Q+R$, is divisible by 9.

---

We need to determine which numbers the balance is divisible by based on this condition.

---

Consider the divisibility rule for 9: A number is divisible by 9 if and only if the sum of its digits is divisible by 9.

Since the sum of the digits $P+Q+R$ is given to be divisible by 9, according to this rule, the number $PQR$ is divisible by 9.

---

Consider the divisibility rule for 3: A number is divisible by 3 if and only if the sum of its digits is divisible by 3.

If the sum of the digits is divisible by 9, it must also be divisible by 3, because 9 is a multiple of 3 ($9 = 3 \times 3$). Any number that is a multiple of 9 is also a multiple of 3.

Since the sum of the digits $P+Q+R$ is divisible by 3 (because it's divisible by 9), according to the divisibility rule for 3, the number $PQR$ is divisible by 3.

---

Therefore, if the sum of the digits $P+Q+R$ is divisible by 9, the number $PQR$ is divisible by both 3 and 9.

---

Comparing this conclusion with the given options:

(A) 3 only - Incorrect.

(B) 9 only - Incorrect.

(C) Both 3 and 9 - Correct.

(D) Neither 3 nor 9 - Incorrect.

---

The correct answer is (C) Both 3 and 9.

Solution:

---

The account balance is a 3-digit number $PQR$, where $P, Q,$ and $R$ are its digits ($P \neq 0$). The value of the number is $100P + 10Q + R$.

The given condition is that the last digit $R$ is an even number.

An even number is an integer that is divisible by 2. The even digits are $0, 2, 4, 6, 8$. So, $R \in \{0, 2, 4, 6, 8\}$.

---

We need to determine which of the given numbers (2, 4, 6, or 8) the balance $PQR$ is divisible by, based on the condition that $R$ is an even digit.

---

Let's apply the divisibility rules:

Divisibility by 2: A number is divisible by 2 if its last digit is an even number (0, 2, 4, 6, or 8).

The problem states that the last digit $R$ is an even number. This is precisely the condition for a number to be divisible by 2.

Therefore, if the last digit $R$ is an even number, the balance $PQR$ is divisible by 2.

---

Let's consider the other options to see if the condition guarantees divisibility by them:

Divisibility by 4: A number is divisible by 4 if the number formed by its last two digits ($QR$) is divisible by 4.

If $R$ is even, the number $QR = 10Q + R$. Is this always divisible by 4? Not necessarily. For example, if $Q=1$ and $R=2$, the number is $12$, which is divisible by 4. But if $Q=1$ and $R=6$, the number is $16$, which is divisible by 4. If $Q=2$ and $R=2$, the number is $22$, which is not divisible by 4. The condition that $R$ is even does not guarantee that $10Q+R$ is divisible by 4 for any digit $Q$. Thus, the number $PQR$ is not guaranteed to be divisible by 4.

Divisibility by 6: A number is divisible by 6 if it is divisible by both 2 and 3.

The condition ($R$ is even) guarantees divisibility by 2. However, it does not guarantee divisibility by 3 (which requires the sum of digits $P+Q+R$ to be divisible by 3). For example, if the number is 102 ($P=1, Q=0, R=2$), the last digit is even (2), so it's divisible by 2. The sum of digits is $1+0+2=3$, which is divisible by 3. So 102 is divisible by 6. But if the number is 104 ($P=1, Q=0, R=4$), the last digit is even (4), so it's divisible by 2. The sum of digits is $1+0+4=5$, which is not divisible by 3. So 104 is not divisible by 6. Thus, the number $PQR$ is not guaranteed to be divisible by 6 just because $R$ is even.

Divisibility by 8: A number is divisible by 8 if the number formed by its last three digits ($PQR$ itself, for a 3-digit number) is divisible by 8. More generally, for any number, it's divisible by 8 if the number formed by its last three digits is divisible by 8.

The condition that the last digit $R$ is even does not guarantee divisibility by 8. For example, 102 has an even last digit (2), but is not divisible by 8 ($102 = 8 \times 12 + 6$). 104 has an even last digit (4), but is not divisible by 8 ($104 = 8 \times 13$). Thus, the number $PQR$ is not guaranteed to be divisible by 8.

---

Based on the divisibility rules, the condition that the last digit $R$ is an even number guarantees only that the number is divisible by 2.

---

Comparing with the given options:

(A) 2 - Correct.

(B) 4 - Not guaranteed.

(C) 6 - Not guaranteed.

(D) 8 - Not guaranteed.

---

The correct answer is (A) 2.

Solution:

---

The problem describes a division where a single-digit number, let's call it $A$, is divided by $5$, and the remainder is $3$.

---

According to the division algorithm, for any integer dividend $A$ and any positive integer divisor $d$, there exist unique integers quotient $q$ and remainder $r$ such that:

$A = d \times q + r$

where $0 \leq r < d$.

---

In this problem, the dividend is $A$, the divisor is $d=5$, and the remainder is $r=3$.

Substituting these values into the division algorithm formula, we get:

$A = 5 \times q + 3$

---

The problem states that $A$ is a single digit. This means $A$ must be an integer between $0$ and $9$, inclusive ($0 \leq A \leq 9$).

Also, the quotient $q$ must be a non-negative integer ($q \geq 0$), and the remainder $r=3$ must satisfy $0 \leq 3 < 5$, which is true.

---

We need to find the value(s) of $A$ by trying different non-negative integer values for the quotient $q$ in equation (i) and checking if the resulting $A$ is a single digit.

---

Case 1: Let $q = 0$

Substitute $q=0$ into equation (i):

$A = 5(0) + 3$

$A = 0 + 3$

$A = 3$

$3$ is a single digit ($0 \leq 3 \leq 9$). So, $A=3$ is a possible value.

---

Case 2: Let $q = 1$

Substitute $q=1$ into equation (i):

$A = 5(1) + 3$

$A = 5 + 3$

$A = 8$

$8$ is a single digit ($0 \leq 8 \leq 9$). So, $A=8$ is a possible value.

---

Case 3: Let $q = 2$

Substitute $q=2$ into equation (i):

$A = 5(2) + 3$

$A = 10 + 3$

$A = 13$

$13$ is not a single digit. So, $A=13$ is not a valid value based on the problem's constraint.

---

For any integer $q \geq 2$, the value of $5q + 3$ will be $13$ or greater, which are not single digits.

Therefore, the only possible single-digit values for $A$ are $3$ and $8$.

---

Comparing the possible values of $A$ with the given options:

(A) 3 - Is a possible value.

(B) 8 - Is a possible value.

(C) Either 3 or 8 - This option covers both possible values.

(D) Cannot be determined - Incorrect, as we found the possible values.

---

The possible value(s) of $A$ is either 3 or 8.

---

The correct answer is (C) Either 3 or 8.

Solution:

The given number is $509$.

---

The general form of a number is expressing it as the sum of the products of its digits with their respective place values.

The number $509$ is a 3-digit number. The digits and their place values are:

Digit $5$ is in the hundreds place (place value $100$).

Digit $0$ is in the tens place (place value $10$).

Digit $9$ is in the units place (place value $1$).

---

To write the number $509$ in the general form, we multiply each digit by its place value and sum the results:

$5 \times \text{(Place value of hundreds)} + 0 \times \text{(Place value of tens)} + 9 \times \text{(Place value of units)}$

$5 \times 100 + 0 \times 10 + 9 \times 1$

---

Let's evaluate this expression:

$5 \times 100 = 500$

$0 \times 10 = 0$

$9 \times 1 = 9$

Sum = $500 + 0 + 9 = 509$. This is the expanded form of the number.

---

Now, let's compare the general form and the expanded form with the given options:

(A) $5 \times 100 + 0 \times 10 + 9 \times 1$. This is the correct representation of the number in general form.

(B) $500 + 9$. This is the expanded form of the number, which is equivalent to the general form when the products are calculated ($5 \times 100 = 500$ and $0 \times 10 + 9 \times 1 = 9$).

(C) $5 \times 10 + 9 \times 1 = 50 + 9 = 59$. This is not the number $509$. This would be the general form for the number $59$.

(D) Both (A) and (B). Since both option (A) (general form) and option (B) (expanded form, directly derived from the general form) correctly represent the number $509$, this option is the most accurate.

---

Thus, both expressions in options (A) and (B) are valid ways to represent the number $509$ based on its place values.

---

The correct answer is (D) Both (A) and (B).

Solution:

---

Let the 2-digit number be $XY$, where $X$ is the tens digit and $Y$ is the units digit. $X$ and $Y$ are single digits, with $X \neq 0$ (for it to be a 2-digit number).

The value of the number $XY$ in general form is $10 \times X + 1 \times Y = 10X + Y$.

---

The reversed number is $YX$, where $Y$ is the tens digit and $X$ is the units digit. $Y$ can be 0 here.

The value of the number $YX$ in general form is $10 \times Y + 1 \times X = 10Y + X$.

---

We need to find the sum of these two numbers:

Sum = $XY + YX = (10X + Y) + (10Y + X)$

---

Combine like terms:

Sum = $(10X + X) + (Y + 10Y)$

Sum = $11X + 11Y$

---

Factor out the common factor $11$:

Sum = $11(X + Y)$

---

The sum of the two numbers is $11(X + Y)$. This expression shows that the sum is always a multiple of $11$.

Therefore, the sum $XY + YX$ is always divisible by $11$.

---

Let's check an example. Let the number be 38 ($X=3, Y=8$).

Original number = 38.

Reversed number = 83.

Sum = $38 + 83 = 121$.

Using the formula $11(X+Y)$: $11(3+8) = 11(11) = 121$.

The number $121$ is divisible by $11$ ($121 \div 11 = 11$).

Consider another example. Let the number be 50 ($X=5, Y=0$).

Original number = 50.

Reversed number = 05 (which is 5).

Sum = $50 + 05 = 55$.

Using the formula $11(X+Y)$: $11(5+0) = 11(5) = 55$.

The number $55$ is divisible by $11$ ($55 \div 11 = 5$).

---

Based on the general form, the sum is always divisible by $11$. Let's check the options.

(A) $X + Y$: The sum is $11(X+Y)$, which is indeed divisible by $X+Y$ (provided $X+Y \neq 0$). However, 11 is a constant factor always present, regardless of the digits, whereas $X+Y$ varies. The question asks what it is *always* divisible by among the given options. 11 is a guaranteed divisor.

(B) $X - Y$: Not necessarily (e.g., sum is 55 for $X=5, Y=0$, $X-Y=5$. 55 is divisible by 5. But for $X=3, Y=8$, $X-Y=-5$. 121 is not divisible by -5).

(C) 11: Correct, as the sum is $11(X+Y)$.

(D) 9: Not necessarily (e.g., sum is 55, not divisible by 9).

---

While the sum is divisible by $X+Y$, option (C) provides a universal divisor (11) that holds for any 2-digit number, which is a stronger and more appropriate answer in this context.

---

The correct answer is (C) 11.

Solution:

---

The sentence to complete is: "A number is divisible by 8 if the number formed by its last _________ digits is divisible by 8."

---

This statement describes the divisibility rule for 8.

The divisibility rule for 8 states that a number is divisible by 8 if and only if the number formed by its last three digits is divisible by 8.

---

This rule is based on the fact that $1000$ is divisible by $8$ ($1000 \div 8 = 125$). Any number can be written in the form $1000k + L$, where $L$ is the number formed by the last three digits (or the number itself if it has less than three digits). Since $1000k$ is always divisible by $8$, the divisibility of the entire number by $8$ depends solely on whether $L$ is divisible by $8$.

---

Let's consider an example. The number $45120$. The last three digits form the number $120$. Is $120$ divisible by $8$? Yes, $120 \div 8 = 15$. Since $120$ is divisible by $8$, the number $45120$ is divisible by $8$. $45120 \div 8 = 5640$.

Consider the number $17234$. The last three digits form the number $234$. Is $234$ divisible by $8$? $234 = 8 \times 29 + 2$. Since $234$ is not divisible by $8$, the number $17234$ is not divisible by $8$.

---

The sentence accurately describes the divisibility rule for 8 when the blank is filled with the word "three".

The complete sentence is: "A number is divisible by 8 if the number formed by its last three digits is divisible by 8."

---

Comparing this with the given options:

(A) One - This is the rule for divisibility by 2 (last digit is even) or 5 or 10 (last digit is 0 or 5, last digit is 0).

(B) Two - This is the rule for divisibility by 4 (last two digits form a number divisible by 4) or 25 (last two digits form a number divisible by 25).

(C) Three - This is the rule for divisibility by 8.

(D) Four - There isn't a standard divisibility rule based on the last four digits for commonly tested numbers.

---

The correct answer is three.

---

The correct answer is (C) Three.

Solution:

---

The given subtraction puzzle is $C B A - A B C = 1 9 8$.

Here, $CBA$ and $ABC$ represent 3-digit numbers, where $A$, $B$, and $C$ are digits (integers from $0$ to $9$).

For $CBA$ and $ABC$ to be 3-digit numbers, the first digit of each number must be non-zero. So, $C \neq 0$ and $A \neq 0$.

We are also given the condition that $C > A$.

---

Let's represent the numbers $CBA$ and $ABC$ in their general form based on their place values:

The number $CBA$ has digit $C$ in the hundreds place, $B$ in the tens place, and $A$ in the units place. Its value is $100 \times C + 10 \times B + 1 \times A = 100C + 10B + A$.

The number $ABC$ has digit $A$ in the hundreds place, $B$ in the tens place, and $C$ in the units place. Its value is $100 \times A + 10 \times B + 1 \times C = 100A + 10B + C$.

---

The given subtraction puzzle can be written as an algebraic equation:

$(100C + 10B + A) - (100A + 10B + C) = 198$

---

Remove the parentheses and distribute the negative sign:

$100C + 10B + A - 100A - 10B - C = 198$

---

Group the terms involving $A$, $B$, and $C$:

$(100C - C) + (10B - 10B) + (A - 100A) = 198$

Simplify the terms:

$99C + 0 + (-99A) = 198$

$99C - 99A = 198$

---

Factor out the common factor $99$ from the left side of the equation:

$99(C - A) = 198$

---

To find the value of $C - A$, divide both sides of the equation by $99$:

$C - A = \frac{198}{99}$

Performing the division:

$198 \div 99 = 2$

So, we have:

$C - A = 2$

---

This result satisfies the given condition that $C > A$, since $C = A + 2$ means $C$ is always greater than $A$. Also, as $A$ and $C$ are digits ($A, C \in \{1, 2, ..., 9\}$ and $C=A+2$), pairs like $(A,C) = (1,3), (2,4), ..., (7,9)$ are possible, ensuring $A \neq 0$ and $C \neq 0$. The value of digit $B$ can be any digit from $0$ to $9$ and does not affect the difference $CBA - ABC$.

---

The value of $C - A$ is $2$.

---

Comparing our result with the given options:

(A) 1

(B) 2

(C) 3

(D) 4

The value $2$ matches option (B).

---

The correct answer is (B) 2.

Solution:

---

We need to evaluate the truthfulness of Assertion (A) and Reason (R) and determine if R correctly explains A.

---

Assertion (A): The number 248 is divisible by 8.

To check if 248 is divisible by 8, we can perform the division:

$248 \div 8 = 31$

Since the division results in an integer with a remainder of 0, the number 248 is divisible by 8.

Alternatively, we can use the divisibility rule for 8.

Therefore, Assertion (A) is True.

---

Reason (R): The number formed by the last three digits is 248, and 248 is divisible by 8 ($248 = 8 \times 31$).

The first part of the reason states the divisibility rule for 8, which for a 3-digit number is that the number itself must be divisible by 8. For numbers with more than three digits, the rule is that the number formed by the last three digits must be divisible by 8. For the number 248 (a 3-digit number), the number formed by its last three digits is indeed 248 itself. So, the statement "The number formed by the last three digits is 248" is correct in this context.

The second part of the reason states that 248 is divisible by 8 and provides the calculation $248 = 8 \times 31$. This calculation is correct, showing that 248 is indeed divisible by 8.

The divisibility rule for 8 is that a number is divisible by 8 if and only if the number formed by its last three digits is divisible by 8. Reason (R) correctly applies this rule to the number 248 and verifies that the condition (divisibility of the last three digits, which form the number 248 itself) is met.

Therefore, Reason (R) is True.

---

Now, let's determine if Reason (R) is the correct explanation for Assertion (A).

Assertion (A) claims that 248 is divisible by 8. Reason (R) explains this by stating the relevant divisibility rule for 8 and showing that 248 satisfies this rule (by showing that the number formed by the last three digits, 248 itself, is divisible by 8).

Reason (R) provides the correct rule and verification based on that rule, which is the standard way to explain the divisibility of 248 by 8 without performing long division.

Therefore, R is the correct explanation of A.

---

Based on our analysis:

Assertion (A) is True.

Reason (R) is True.

Reason (R) is the correct explanation of Assertion (A).

---

Comparing this with the given options:

(A) Both A and R are true, and R is the correct explanation of A. This matches our conclusion.

(B) Both A and R are true, but R is not the correct explanation of A.

(C) A is true, but R is false.

(D) A is false, but R is true.

---

The correct answer is (A) Both A and R are true, and R is the correct explanation of A.

Solution:

---

The given number is $9876A$, which is a five-digit number where $A$ represents the digit in the units place.

The problem states that the number $9876A$ is divisible by $10$.

---

According to the divisibility rule for $10$, a number is divisible by $10$ if and only if its last digit is $0$.

---

In the number $9876A$, the last digit is $A$.

For the number $9876A$ to be divisible by $10$, its last digit $A$ must be $0$.

Since $A$ is a digit, it can be any integer from $0$ to $9$. The condition that $A$ must be $0$ means that $A$ can only take the value $0$.

---

Let's verify this with an example. If $A=0$, the number is $98760$. $98760 \div 10 = 9876$. So, $98760$ is divisible by $10$.

If we take any other digit for $A$, say $A=5$, the number is $98765$. $98765 \div 10$ leaves a remainder of $5$. So, $98765$ is not divisible by $10$.

---

Based on the divisibility rule for $10$, the digit $A$ must be $0$.

---

Comparing this conclusion with the given options:

(A) Any even digit - Incorrect, only 0 works.

(B) 0 or 5 - Incorrect, only 0 works.

(C) 0 - Correct.

(D) Any digit - Incorrect.

---

The correct answer is (C) 0.

Solution:

---

The problem asks for the general form of a 2-digit number with a specified units digit and tens digit.

The units digit is given as $y$. The place value of the units digit is $1$.

The tens digit is given as $x$. The place value of the tens digit is $10$.

---

The general form of a number is the sum of the products of each digit and its corresponding place value.

For a 2-digit number with tens digit $x$ and units digit $y$, the value of the number is:

$( \text{Tens digit} \times \text{Place value of tens} ) + ( \text{Units digit} \times \text{Place value of units} )$

$(x \times 10) + (y \times 1)$

This simplifies to:

$10x + y$

---

Let's consider an example. Let the number be 47. The tens digit is 4, so $x=4$. The units digit is 7, so $y=7$.

Using the general form $10x + y$: $10(4) + 7 = 40 + 7 = 47$. This correctly represents the number.

---

Now, let's examine the given options:

(A) $xy$ - This usually represents the product of $x$ and $y$, not the number formed by placing the digits $x$ and $y$ next to each other. For example, if $x=4$ and $y=7$, $xy$ would be $4 \times 7 = 28$, not $47$.

(B) $10x + y$ - This matches the general form we derived.

(C) $10y + x$ - This would be the general form if $y$ were the tens digit and $x$ were the units digit (i.e., the number $yx$). For $x=4, y=7$, this gives $10(7) + 4 = 70 + 4 = 74$, which is the reversed number, not the original number $47$.

(D) $x+y$ - This is the sum of the digits, not the value of the number. For $x=4, y=7$, this gives $4+7=11$, not $47$.

---

Therefore, the correct general form of a 2-digit number whose units digit is $y$ and tens digit is $x$ is $10x + y$.

---

The correct answer is (B) $10x + y$.

Solution:

---

The given multiplication puzzle is: $1 2 A \times 5 = 6 B 0$.

Here, $12A$ represents a three-digit number where $A$ is the units digit. $6B0$ represents a three-digit number where $B$ is the tens digit. $A$ and $B$ are single digits, meaning they are integers from $0$ to $9$.

---

We analyze the multiplication in columns, starting from the units column.

In the units column, we multiply the units digit of $12A$, which is $A$, by $5$. The result must have a units digit of $0$ (as seen in the product $6B0$).

We need to find the single digit values for $A$ ($0 \leq A \leq 9$) such that the product $A \times 5$ ends with the digit $0$.

---

Let's check the product $A \times 5$ for each possible single digit value of $A$:

If $A = 0$, $0 \times 5 = 0$. The units digit is $0$.

If $A = 1$, $1 \times 5 = 5$. The units digit is $5$.

If $A = 2$, $2 \times 5 = 10$. The units digit is $0$.

If $A = 3$, $3 \times 5 = 15$. The units digit is $5$.

If $A = 4$, $4 \times 5 = 20$. The units digit is $0$.

If $A = 5$, $5 \times 5 = 25$. The units digit is $5$.

If $A = 6$, $6 \times 5 = 30$. The units digit is $0$.

If $A = 7$, $7 \times 5 = 35$. The units digit is $5$.

If $A = 8$, $8 \times 5 = 40$. The units digit is $0$.

If $A = 9$, $9 \times 5 = 45$. The units digit is $5$.

---

The single-digit values of $A$ for which $A \times 5$ ends in $0$ are $0, 2, 4, 6, 8$.

These are the possible values for $A$ based solely on the units column analysis.

---

Now, let's examine the given options to see which one lists possible values for $A$ from this set $\{0, 2, 4, 6, 8\}$:

(A) 0 or 5: $0$ is possible, but $5$ is not possible based on the units digit ending in $0$.

(B) 0 or 2: $0$ is possible, and $2$ is possible. This option lists two possible values.

(C) 0 or 10: $0$ is possible, but $10$ is not a single digit ($A$ must be a single digit).

(D) Only 0: $0$ is possible, but this option implies that $0$ is the *only* possible value, which is incorrect as $2, 4, 6, 8$ are also possible values for $A$ from the units column analysis.

---

Among the given options, option (B) correctly lists two values ($0$ and $2$) that are indeed possible values for $A$ based on the units column multiplication ending in $0$. While there are other possible values ($4, 6, 8$), option (B) contains only valid possibilities from the context of the units column analysis and is the best fit among the choices provided.

---

The correct answer is (B) 0 or 2.

Solution:

---

The multiplication puzzle is given as $1 2 A \times 5 = 6 B 0$.

Here, $12A$ represents a 3-digit number where $A$ is the units digit, and $6B0$ represents a 3-digit number where $B$ is the tens digit.

The number $12A$ can be written as $100 \times 1 + 10 \times 2 + A = 120 + A$.

The number $6B0$ can be written as $100 \times 6 + 10 \times B + 0 = 600 + 10B$.

---

The puzzle can be written as an equation:

$(120 + A) \times 5 = 600 + 10B$

---

The question asks whether the case $A=0$ satisfies the puzzle.

Let's substitute $A=0$ into the equation:

$(120 + 0) \times 5 = 600 + 10B$

$120 \times 5 = 600 + 10B$

Calculate the product on the left side:

$120 \times 5 = 600$

So the equation becomes:

$600 = 600 + 10B$

---

Now, solve for $B$:

$600 - 600 = 10B$

$0 = 10B$

$B = \frac{0}{10}$

$B = 0$

---

So, if $A=0$, the puzzle implies that $B$ must also be $0$. Both $A=0$ and $B=0$ are valid single digits ($0 \leq 0 \leq 9$).

Substituting these values back into the original puzzle format $12A \times 5 = 6B0$:

$120 \times 5 = 600$

This is a true statement, and the structure fits the form $12A \times 5 = 6B0$ with $A=0$ and $B=0$.

---

Therefore, $A=0$ along with the resulting $B=0$ satisfies the puzzle.

---

Let's evaluate the given options:

(A) Yes - This aligns with our conclusion.

(B) No, because B must be a different digit from A. - The problem statement does not impose the condition that $A$ and $B$ must be different digits. They can be the same.

(C) No, the product 600 does not fit the form 6B0. - This is incorrect. The form $6B0$ represents a number with digit $6$ in the hundreds place, digit $B$ in the tens place, and digit $0$ in the units place. When $B=0$, this number is $600$. So, $600$ fits the form $6B0$ when $B=0$.

(D) Cannot determine from A=0 alone. - We were able to determine the value of $B$ when $A=0$ and verify that it satisfies the puzzle.

---

The case $A=0$ results in $B=0$, and the equation $120 \times 5 = 600$ is true, fitting the puzzle structure. So, $A=0$ is a possible value for $A$ that satisfies the puzzle.

---

The correct answer is (A) Yes.

Solution:

---

The multiplication puzzle is $1 2 A \times 5 = 6 B 0$.

The number $12A$ has units digit $A$. The product $6B0$ has units digit $0$.

---

The question specifically examines the case where $A=5$. Let's perform the multiplication with $A=5$:

Substitute $A=5$ into $12A \times 5$: $(120 + 5) \times 5 = 125 \times 5$.

Calculate the product $125 \times 5$: $125 \times 5 = 625$.

---

The puzzle states that the product should be of the form $6B0$. This means the product must be a 3-digit number starting with 6 and ending with 0.

The product we obtained when $A=5$ is $625$.

---

Comparing $625$ with the required form $6B0$:

The hundreds digit is 6, which matches.

The units digit of the product $625$ is $5$.

The units digit required by the form $6B0$ is $0$.

Since the units digit of $625$ ($5$) does not match the required units digit of $6B0$ ($0$), the product $625$ does not fit the form $6B0$.

---

Therefore, $A=5$ does not satisfy the puzzle.

---

Let's evaluate the given options:

(A) Yes, because $125 \times 5 = 625$ which matches the form $6B0$ with B=2. - This is incorrect. $625$ does not match the form $6B0$ because their units digits are different (5 vs 0).

(B) No, because the units digit of the product should be 0, not 5. - This is correct. The product must end in 0 as specified by the form $6B0$, and $125 \times 5$ ends in 5.

(C) Yes, provided B can be 5. - This is irrelevant. The units digit must be 0, regardless of the value of B.

(D) No, because B must be different from A. - The problem does not require B to be different from A.

---

The reason $A=5$ does not satisfy the puzzle is that the resulting product does not have a units digit of 0.

---

The correct answer is (B) No, because the units digit of the product should be 0, not 5.

Solution:

---

The puzzle is $1 2 A \times 5 = 6 B 0$, where $A$ and $B$ are single digits ($0 \leq A \leq 9$, $0 \leq B \leq 9$).

The number $12A$ can be written as $120 + A$. The number $6B0$ can be written as $600 + 10B$.

The puzzle is equivalent to the equation:

$(120 + A) \times 5 = 600 + 10B$

$120 \times 5 + A \times 5 = 600 + 10B$

$600 + 5A = 600 + 10B$

---

Subtracting $600$ from both sides gives:

$5A = 10B$

---

Dividing both sides by $5$ gives the relationship between $A$ and $B$:

$A = 2B$

---

We also know from the units column of the multiplication ($A \times 5$ ends in 0) that the units digit of $A \times 5$ must be 0. This happens when $A$ is a multiple of 2 (since $5 \times \text{even number}$ ends in 0) or when $A=0$ (since $0 \times 5 = 0$) or when $A=5$ (since $5 \times 5 = 25$ ends in 5).

More rigorously, $A \times 5$ must end in 0. Possible single digits $A$ are $0, 1, 2, 3, 4, 5, 6, 7, 8, 9$.

Products $A \times 5$: $0, 5, 10, 15, 20, 25, 30, 35, 40, 45$.

The units digit of the product is 0 when $A \in \{0, 2, 4, 6, 8\}$.

---

The preamble of the question incorrectly states "The possible digits for A are 0 or 5" based on the units column. Based on the units column ($A \times 5$ ending in 0 in $6B0$), the possible values for $A$ are $\{0, 2, 4, 6, 8\}$. The preamble then correctly eliminates $A=5$ but incorrectly concludes $A=0$ is the only possibility.

---

We must satisfy the equation $A = 2B$ where $A \in \{0, 2, 4, 6, 8\}$ and $B \in \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$.

Let's find pairs $(A, B)$ that satisfy $A=2B$ with $B$ being a single digit:

• If $B=0$, $A = 2(0) = 0$. $(A,B) = (0,0)$. $A=0$ is in $\{0,2,4,6,8\}$. This is a valid solution.
• If $B=1$, $A = 2(1) = 2$. $(A,B) = (2,1)$. $A=2$ is in $\{0,2,4,6,8\}$. This is a valid solution.
• If $B=2$, $A = 2(2) = 4$. $(A,B) = (4,2)$. $A=4$ is in $\{0,2,4,6,8\}$. This is a valid solution.
• If $B=3$, $A = 2(3) = 6$. $(A,B) = (6,3)$. $A=6$ is in $\{0,2,4,6,8\}$. This is a valid solution.
• If $B=4$, $A = 2(4) = 8$. $(A,B) = (8,4)$. $A=8$ is in $\{0,2,4,6,8\}$. This is a valid solution.
• If $B=5$, $A = 2(5) = 10$. $10$ is not a single digit. Stop.

So, the possible solutions $(A,B)$ are $(0,0), (2,1), (4,2), (6,3), (8,4)$.

---

The question asks, "if A and B must be distinct, is there a valid solution?".

We need to check which of the valid solutions have $A \neq B$:

• $(0,0)$: $A=B$. Not distinct.
• $(2,1)$: $A \neq B$. Distinct. This is a valid solution with distinct digits. ($122 \times 5 = 610$)
• $(4,2)$: $A \neq B$. Distinct. This is a valid solution with distinct digits. ($124 \times 5 = 620$)
• $(6,3)$: $A \neq B$. Distinct. This is a valid solution with distinct digits. ($126 \times 5 = 630$)
• $(8,4)$: $A \neq B$. Distinct. This is a valid solution with distinct digits. ($128 \times 5 = 640$)

---

Since we found solutions where $A$ and $B$ are distinct digits (e.g., $(2,1)$, $(4,2)$, etc.), the answer to the question "is there a valid solution [with distinct digits]?" is Yes.

Let's look at the options:

(A) Yes, A=0, B=0 is a valid solution. - This is true, but $A=B$, so it's not a solution with distinct digits. This doesn't directly answer the question asked.

(B) Yes, there is another solution with distinct digits. - This is true. There are multiple solutions with distinct digits (e.g., (2,1), (4,2), etc.). The word "another" implies a solution different from the non-distinct one (0,0), which fits.

(C) No, there is no solution with distinct digits for A and B. - This is false.

(D) The problem is flawed if A and B must be distinct. - This is false, as distinct solutions exist.

---

Therefore, there exist solutions to the puzzle where the digits $A$ and $B$ are distinct.

---

The correct answer is (B) Yes, there is another solution with distinct digits.

Solution:

---

We need to determine which of the given numbers is divisible by $11$.

---

The divisibility rule for $11$ states that a number is divisible by $11$ if the alternating sum of its digits, starting from the rightmost digit with a plus sign, is divisible by $11$.

---

Let's apply this rule to each option:

(A) 121:

Digits from right to left: 1, 2, 1.

Alternating sum = $+1 - 2 + 1 = 0$.

Since $0$ is divisible by $11$ ($0 \div 11 = 0$), the number $121$ is divisible by $11$.

---

(B) 1331:

Digits from right to left: 1, 3, 3, 1.

Alternating sum = $+1 - 3 + 3 - 1 = 0$.

Since $0$ is divisible by $11$ ($0 \div 11 = 0$), the number $1331$ is divisible by $11$.

---

(C) 24684:

Digits from right to left: 4, 8, 6, 4, 2.

Alternating sum = $+4 - 8 + 6 - 4 + 2 = (4 + 6 + 2) - (8 + 4) = 12 - 12 = 0$.

Since $0$ is divisible by $11$ ($0 \div 11 = 0$), the number $24684$ is divisible by $11$.

---

(D) 10813:

Digits from right to left: 3, 1, 8, 0, 1.

Alternating sum = $+3 - 1 + 8 - 0 + 1 = (3 + 8 + 1) - (1 + 0) = 12 - 1 = 11$.

Since $11$ is divisible by $11$ ($11 \div 11 = 1$), the number $10813$ is divisible by $11$.

---

Based on the divisibility rule for $11$, all the given numbers $121, 1331, 24684,$ and $10813$ are divisible by $11$.

In a single-choice question format, there is typically only one correct answer. However, applying the standard divisibility rule shows that all options are divisible by 11. Assuming there might be an error in the question or options provided in the input, we present the verification for each.

---

Given that the question asks "Which of the following numbers is divisible by 11?" and all listed options satisfy this condition, there seems to be an inconsistency in the question format expecting a single answer. However, option (A) is the first number listed that is divisible by 11.

---

The correct answer is (A) 121 (while noting that options B, C, and D are also divisible by 11).

Solution:

---

The problem asks us to complete a sentence about the general form of a 3-digit number $cba$.

The number is given as $cba$, where $c$ is the digit in the hundreds place, $b$ is the digit in the tens place, and $a$ is the digit in the units place.

The general form of this number is given as $100c + 10b + a$.

---

For a number to be a 3-digit number, the digit in the highest place value position (the hundreds place in this case) cannot be zero.

The digit in the hundreds place of the number $cba$ is $c$.

Therefore, for $cba$ to be a 3-digit number, the digit $c$ must be a non-zero digit.

In other words, $c$ cannot be equal to $0$.

$c \neq 0$

---

The sentence provided is: "The general form of a 3-digit number $cba$ is $100c + 10b + a$, where $a, b, c$ are the digits at the units, tens, and hundreds place respectively, and $c \neq$ _________."

Based on the definition of a 3-digit number, the blank should be filled with the digit that the hundreds place digit ($c$) cannot be.

That digit is $0$.

---

Let's check the options:

(A) 1 - If $c=1$, the number is $1ba$, which is a valid 3-digit number (e.g., 123). So $c$ can be 1.

(B) 0 - If $c=0$, the number would be $0ba$, which is a 2-digit number ($10b+a$), not a 3-digit number (unless $b=0$ and $a=0$, which is just the number 0). So $c$ cannot be 0 for it to be a 3-digit number.

(C) 9 - If $c=9$, the number is $9ba$, which is a valid 3-digit number (e.g., 987). So $c$ can be 9.

(D) Any even number - $c$ can be even numbers like 2, 4, 6, 8 (e.g., 215, 450, 601, 888), and these are valid 3-digit numbers. The digit 0 is also even, and $c$ cannot be 0. But the condition is specifically about the leading digit not being 0, regardless of whether it's even or odd.

---

The condition for $cba$ to be a 3-digit number is that $c$ must be a digit from 1 to 9. This is equivalent to stating that $c$ must be a digit and $c \neq 0$.

---

The correct answer to complete the sentence is 0.

---

The correct answer is (B) 0.

Solution:

---

The given number is $5A$, which is a 2-digit number. Here, $5$ is the tens digit and $A$ is the units digit. $A$ must be a single digit, i.e., $0 \leq A \leq 9$.

The problem states that the number $5A$ is divisible by $3$.

---

According to the divisibility rule for $3$, a number is divisible by $3$ if and only if the sum of its digits is divisible by $3$.

---

The digits of the number $5A$ are $5$ and $A$.

The sum of the digits is $5 + A$.

---

For the number $5A$ to be divisible by $3$, the sum of its digits $(5 + A)$ must be divisible by $3$.

We need to find the single digit values for $A$ ($0 \leq A \leq 9$) such that $5 + A$ is a multiple of $3$.

---

Let's check the sum $5 + A$ for possible values of $A$ and see if it is divisible by 3:

• If $A=0$, sum = $5 + 0 = 5$. $5 \div 3 = 1$ remainder $2$. Not divisible by 3.
• If $A=1$, sum = $5 + 1 = 6$. $6 \div 3 = 2$. Divisible by 3. So $A=1$ is possible.
• If $A=2$, sum = $5 + 2 = 7$. $7 \div 3 = 2$ remainder $1$. Not divisible by 3.
• If $A=3$, sum = $5 + 3 = 8$. $8 \div 3 = 2$ remainder $2$. Not divisible by 3.
• If $A=4$, sum = $5 + 4 = 9$. $9 \div 3 = 3$. Divisible by 3. So $A=4$ is possible.
• If $A=5$, sum = $5 + 5 = 10$. $10 \div 3 = 3$ remainder $1$. Not divisible by 3.
• If $A=6$, sum = $5 + 6 = 11$. $11 \div 3 = 3$ remainder $2$. Not divisible by 3.
• If $A=7$, sum = $5 + 7 = 12$. $12 \div 3 = 4$. Divisible by 3. So $A=7$ is possible.
• If $A=8$, sum = $5 + 8 = 13$. $13 \div 3 = 4$ remainder $1$. Not divisible by 3.
• If $A=9$, sum = $5 + 9 = 14$. $14 \div 3 = 4$ remainder $2$. Not divisible by 3.

---

The single digit values of $A$ for which the sum $5+A$ is divisible by 3 are $1, 4,$ and $7$.

---

The possible values for $A$ are 1 or 4 or 7.

---

Let's examine the given options:

(A) 1 or 4 or 7 - This lists the exact possible values we found.

(B) 0 or 3 or 6 or 9 - These values do not make $5+A$ divisible by 3.

(C) 1, 4, 7 (sum of digits 5+A must be divisible by 3) - This option correctly identifies the possible values and provides the correct reasoning.

(D) 2, 5, 8 (sum of digits 5+A must be divisible by 3) - These values do not make $5+A$ divisible by 3.

---

Options (A) and (C) both list the correct possible values. Option (C) provides the correct reasoning, which is helpful context, but option (A) directly answers the question about the possible value(s) of A.

In a standard multiple-choice question, both (A) and (C) might be considered correct depending on the question setter's intent. However, option (C) gives more information justifying the answer. Let's choose the option that simply lists the values.

---

The correct answer is (A) 1 or 4 or 7.

The 2-digit number is $47$.

In a 2-digit number, the first digit from the left is the tens digit and the second digit is the units digit.

For the number $47$, the tens digit is $4$ and the units digit is $7$.

The general form of a 2-digit number with tens digit $a$ and units digit $b$ is $10a + b$.

To write the number $47$ in its general form using digits $a$ and $b$, we let the tens digit be $a=4$ and the units digit be $b=7$.

So, the general form of $47$ is:

$10 \times (\text{tens digit}) + (\text{units digit})$

$10 \times 4 + 7$

Which can be written as $10a + b$ where $a=4$ and $b=7$.

---

Now, we need to write the general form of a 2-digit number with units digit $a$ and tens digit $b$.

The tens digit is given as $b$ and the units digit is given as $a$.

Using the formula for the general form of a 2-digit number:

$10 \times (\text{tens digit}) + (\text{units digit})$

Substitute the given digits:

$10 \times b + a$

Thus, the general form of a 2-digit number with units digit $a$ and tens digit $b$ is $10b + a$.

The 3-digit number is $582$.

In a 3-digit number, the first digit from the left is the hundreds digit, the second digit is the tens digit, and the third digit is the units digit.

For the number $582$, the hundreds digit is $5$, the tens digit is $8$, and the units digit is $2$.

The general form of a 3-digit number with hundreds digit $a$, tens digit $b$, and units digit $c$ is $100a + 10b + c$.

To write the number $582$ in its general form using digits $a, b,$ and $c$, we let the hundreds digit be $a=5$, the tens digit be $b=8$, and the units digit be $c=2$.

So, the general form of $582$ is:

$100 \times (\text{hundreds digit}) + 10 \times (\text{tens digit}) + (\text{units digit})$

$100 \times 5 + 10 \times 8 + 2$

Which can be written as $100a + 10b + c$ where $a=5$, $b=8$, and $c=2$.

---

Now, we need to write the general form of a 3-digit number with hundreds digit $a$, tens digit $b$, and units digit $c$.

The hundreds digit is given as $a$, the tens digit is given as $b$, and the units digit is given as $c$.

Using the formula for the general form of a 3-digit number:

$100 \times (\text{hundreds digit}) + 10 \times (\text{tens digit}) + (\text{units digit})$

Substitute the given digits:

$100 \times a + 10 \times b + c$

Thus, the general form of a 3-digit number with hundreds digit $a$, tens digit $b$, and units digit $c$ is $100a + 10b + c$.

The given 2-digit number is in the general form $10a + b$.

In this general form, $a$ represents the tens digit and $b$ represents the units digit.

To obtain the number by reversing the digits, the units digit becomes the new tens digit and the tens digit becomes the new units digit.

So, the new tens digit is $b$ and the new units digit is $a$.

The general form of the number obtained by reversing the digits is $10 \times (\text{new tens digit}) + (\text{new units digit})$.

Substituting the new digits, we get $10 \times b + a$, which is $10b + a$.

---

Now, we need to find the sum of the original number and the reversed number in general form.

Original number = $10a + b$

Reversed number = $10b + a$

Sum = (Original number) + (Reversed number)

Sum = $(10a + b) + (10b + a)$

Combine the terms with $a$ and the terms with $b$:

Sum = $(10a + a) + (b + 10b)$

Sum = $11a + 11b$

We can factor out the common factor, $11$:

Sum = $11(a + b)$

So, the sum of the original number and the reversed number in general form is $11(a + b)$.

The original 2-digit number is given in the general form $10a + b$.

Here, $a$ is the tens digit and $b$ is the units digit.

The number obtained by reversing the digits has $b$ as the tens digit and $a$ as the units digit.

The general form of the reversed number is $10b + a$.

---

We need to find the difference between the original number and the reversed number, given that $a > b$.

Since $a > b$, the original number ($10a + b$) is greater than the reversed number ($10b + a$).

Difference = (Original number) - (Reversed number)

Difference = $(10a + b) - (10b + a)$

Remove the parentheses and change the signs for the terms in the second parenthesis:

Difference = $10a + b - 10b - a$

Group the like terms:

Difference = $(10a - a) + (b - 10b)$

Perform the subtractions:

Difference = $9a - 9b$

Factor out the common factor, $9$:

Difference = $9(a - b)$

Thus, the difference between the original number and the number obtained by reversing its digits, assuming $a > b$, is $9(a - b)$.

Let the 2-digit number be represented in its general form as $10a + b$, where $a$ is the tens digit ($a \neq 0$) and $b$ is the units digit.

The number obtained by reversing the digits has $b$ as the tens digit and $a$ as the units digit. Its general form is $10b + a$.

---

Now, let's find the sum of the original number and the number obtained by reversing its digits:

Sum = (Original number) + (Reversed number)

Sum = $(10a + b) + (10b + a)$

Combine the terms with $a$ and the terms with $b$:

Sum = $(10a + a) + (b + 10b)$

Sum = $11a + 11b$

Factor out the common factor, $11$:

Sum = $11(a + b)$

Since $a$ and $b$ are digits, $a+b$ is an integer. The sum is $11$ multiplied by an integer $(a+b)$.

Therefore, the sum of a 2-digit number and the number obtained by reversing its digits is always divisible by $11$.

---

Example:

Let the 2-digit number be $47$.

Here, the tens digit is $a=4$ and the units digit is $b=7$.

Original number (general form): $10(4) + 7 = 40 + 7 = 47$.

The number obtained by reversing the digits is $74$.

Reversed number (general form): $10(7) + 4 = 70 + 4 = 74$.

Sum of the original number and the reversed number = $47 + 74 = 121$.

Let's check if the sum $121$ is divisible by $11$:

$121 \div 11 = 11$

The sum $121$ is indeed divisible by $11$.

Using the derived general form for the sum: $11(a + b) = 11(4 + 7) = 11(11) = 121$.

This confirms that the sum is always divisible by $11$.

Let the 2-digit number be represented in its general form as $10a + b$, where $a$ is the tens digit ($a$ can be any integer from $1$ to $9$) and $b$ is the units digit ($b$ can be any integer from $0$ to $9$).

The number obtained by reversing the digits has $b$ as the tens digit and $a$ as the units digit. Its general form is $10b + a$.

---

Now, let's find the difference between the original number and the number obtained by reversing its digits.

Difference = (Original number) - (Reversed number)

Difference = $(10a + b) - (10b + a)$

Remove the parentheses and change the signs for the terms in the second parenthesis:

Difference = $10a + b - 10b - a$

Group the like terms:

Difference = $(10a - a) + (b - 10b)$

Perform the subtractions:

Difference = $9a - 9b$

Factor out the common factor, $9$:

Difference = $9(a - b)$

Since $a$ and $b$ are integers representing digits, their difference $(a - b)$ is also an integer. The difference between the original number and the reversed number is $9$ multiplied by an integer $(a-b)$.

Therefore, the difference between a 2-digit number and the number obtained by reversing its digits is always divisible by $9$. If $b > a$, the difference would be $(10b+a) - (10a+b) = 9b - 9a = 9(b-a)$. In either case, the result is a multiple of $9$, so the difference (magnitude) is divisible by $9$.

---

Example:

Let the 2-digit number be $62$.

Here, the tens digit is $a=6$ and the units digit is $b=2$.

Original number (general form): $10(6) + 2 = 60 + 2 = 62$.

The number obtained by reversing the digits is $26$.

Reversed number (general form): $10(2) + 6 = 20 + 6 = 26$.

Difference between the numbers = $62 - 26 = 36$.

Let's check if the difference $36$ is divisible by $9$:

$36 \div 9 = 4$

The difference $36$ is indeed divisible by $9$.

Using the derived general form for the difference: $9(a - b) = 9(6 - 2) = 9(4) = 36$.

This confirms that the difference is always divisible by $9$.

The given addition problem is:

``` 4 A + 9 8 ----- 1 4 5 ```

---

We need to find the value of the digit A.

Let's analyze the addition column by column, starting from the units column.

---

Units Column:

The units digits are A and 8. Their sum results in a number whose units digit is 5.

So, $A + 8$ must end in 5.

Possible values for $A + 8$ are $5, 15, 25, ...$

Since A is a single digit (from 0 to 9), the maximum value of $A + 8$ is $9 + 8 = 17$.

Therefore, $A + 8$ must be $15$.

$A + 8 = 15$

Subtract 8 from both sides:

$A = 15 - 8$

$A = 7$

When $A = 7$, the units column addition is $7 + 8 = 15$. The units digit of the sum is 5, and there is a carry-over of 1 to the tens column.

---

Tens Column:

The tens digits are 4 and 9. We must also add the carry-over from the units column, which is 1.

Sum in tens column = $4 + 9 + (\text{carry-over})$

Sum in tens column = $4 + 9 + 1$

Sum in tens column = $13 + 1$

Sum in tens column = $14$

The result in the sum is 145. This means the tens digit of the sum is 4 and there is a carry-over of 1 to the hundreds column (which appears as the leading 1 in 145).

Our calculation $4 + 9 + 1 = 14$ matches this, with 4 being the tens digit and 1 being carried over to the hundreds place.

---

The value of A that satisfies the units column addition is $A = 7$, and this value is consistent with the tens column addition and the given sum.

---

The value of the digit A is $\mathbf{7}$.

The given addition problem is:

``` A A + A ----- 1 5 0 ```

We are asked to find the values of the digits A and B. Note that the digit B does not appear in the visual representation of the addition problem provided.

Let's interpret the problem as a standard column addition where 'AA' represents a 2-digit number $10A + A$ and 'A' represents a single-digit number $A$. The sum is $150$.

---

We can set up the addition by looking at each column:

``` Tens | Units -----|------- A | A + | A -----|------- 1 | 5 | 0 ```

Units Column:

The units digits being added are $A$ and $A$. The units digit of the sum is $0$. Let $c_1$ be the carry-over to the tens column.

$A + A = 10 \times c_1 + 0$

Since A is a single digit (0-9), the possible values for $2A$ are $0, 2, 4, ..., 18$. For $2A$ to be a multiple of 10, $2A$ must be $0$ or $10$.

If $2A = 0$, then $A = 0$. In this case, $10c_1 = 0$, so $c_1 = 0$.

If $2A = 10$, then $A = 5$. In this case, $10c_1 = 10$, so $c_1 = 1$.

So, from the units column, possible pairs of $(A, c_1)$ are $(0, 0)$ or $(5, 1)$.

---

Tens Column:

The tens digit of the first number ('AA') is $A$. The second number ('A') has no tens digit (effectively 0). We must add the carry-over $c_1$ from the units column. The tens digit of the sum ('150') is $5$. Let $c_2$ be the carry-over to the hundreds column.

$A + 0 + c_1 = 10 \times c_2 + 5$

---

Hundreds Column:

There are no hundreds digits in the numbers being added. The hundreds digit of the sum ('150') is $1$. This digit must be the carry-over $c_2$ from the tens column.

---

Now, substitute the value of $c_2$ into the equation for the tens column:

$A + c_1 = 10(1) + 5$

---

Finally, we test the possible pairs of $(A, c_1)$ derived from the units column in this equation:

Case 1: $A = 0, c_1 = 0$

Substitute these values into $A + c_1 = 15$:

$0 + 0 = 15$

$0 = 15$

This is false. So, $A$ cannot be $0$.

---

Case 2: $A = 5, c_1 = 1$

Substitute these values into $A + c_1 = 15$:

$5 + 1 = 15$

$6 = 15$

This is false. So, $A$ cannot be $5$.

---

Since there are no other possibilities for the single digit $A$ based on the units column, and neither possibility satisfies the conditions from the tens and hundreds columns, there is no solution for the digit A that fits the standard rules of column addition for single digits (0-9) with the given sum 150.

---

Regarding the digit B, it does not appear in the given addition problem image. If the question implies that the sum of 'AA' and 'A' is 150, then the problem as stated has no valid single digit solution for A. It is possible there is a typo in the question or the image.

Based on the provided image and the requirement for A to be a digit, there is no solution.

The given multiplication problem is presented as:

``` A x A --- 9 A ```

In this problem, A represents a single digit. The multiplication is of the single digit A by itself ($A \times A$). The result is a 2-digit number represented as 9A.

In the 2-digit number 9A, the digit 9 is the tens digit and the digit A is the units digit. The value of a 2-digit number with tens digit $t$ and units digit $u$ is $10t + u$.

Therefore, the 2-digit number 9A has the value $10 \times 9 + A = 90 + A$.

---

The multiplication problem can be translated into the following algebraic equation:

$A \times A = 90 + A$

This simplifies to:

$A^2 = 90 + A$

To solve for A, we can rearrange this equation into a standard quadratic form $ax^2 + bx + c = 0$:

$A^2 - A - 90 = 0$

This is a quadratic equation with $a=1$, $b=-1$, and $c=-90$. We can solve for A using the quadratic formula:

$A = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Substitute the values of $a$, $b$, and $c$ into the formula:

$A = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-90)}}{2(1)}$

$A = \frac{1 \pm \sqrt{1 + 360}}{2}$

$A = \frac{1 \pm \sqrt{361}}{2}$

We know that $19 \times 19 = 361$, so $\sqrt{361} = 19$.

$A = \frac{1 \pm 19}{2}$

This gives us two possible values for A:

$A_1 = \frac{1 + 19}{2} = \frac{20}{2} = 10$

$A_2 = \frac{1 - 19}{2} = \frac{-18}{2} = -9$

---

The problem states that A is a digit. By definition, a digit is one of the single symbols used to represent numbers in a positional numeral system. In the standard decimal system, the digits are $0, 1, 2, 3, 4, 5, 6, 7, 8, 9$.

We must check if the values we found for A are valid digits.

The first value, $A_1 = 10$, is not a single digit as it requires two symbols (1 and 0) to represent it.

The second value, $A_2 = -9$, is not a digit as digits are typically non-negative integers from 0 to 9.

Since neither of the solutions for A is a valid single digit from 0 to 9, there is no single digit value for A that satisfies the given multiplication problem under standard mathematical interpretation.

The multiplication problem can be written as an equation in terms of the digits A, B, and C.

The number AB is $10A + B$. The number CA0 is $100C + 10A + 0 = 100C + 10A$.

The equation is: $(10A + B) \times 5 = 100C + 10A$.

---

Expand the equation:

$50A + 5B = 100C + 10A$

Subtract $10A$ from both sides:

$40A + 5B = 100C$

Divide the entire equation by 5:

---

A, B, and C are single digits (0-9). Since 'AB' is a 2-digit number, $A \neq 0$. Since 'CA0' is a 3-digit number, $C \neq 0$. So, $A \in \{1, ..., 9\}$, $B \in \{0, ..., 9\}$, $C \in \{1, ..., 9\}$.

We need to find integer values for A and B within their range that satisfy $8A + B = 20C$ for some integer C within its range.

---

Let's test values for C from 1 to 9:

If $C=1$: $8A + B = 20$. Possible values for A (1-9) and B (0-9). If $A=1$, $8+B=20 \implies B=12$ (not a digit). If $A=2$, $16+B=20 \implies B=4$. This is a valid solution: A=2, B=4 (C=1, check: $24 \times 5 = 120$). For $A \ge 3$, $8A > 20$, so no non-negative B is possible.

If $C=2$: $8A + B = 40$. If $A=1$, $8+B=40 \implies B=32$ (not a digit). ... If $A=4$, $32+B=40 \implies B=8$. Valid solution: A=4, B=8 (C=2, check: $48 \times 5 = 240$). If $A=5$, $40+B=40 \implies B=0$. Valid solution: A=5, B=0 (C=2, check: $50 \times 5 = 250$). For $A \ge 6$, $8A > 40$, so no non-negative B is possible.

If $C=3$: $8A + B = 60$. If $A=1..6$, $8A+B < 60$ for $B \le 9$. If $A=7$, $56+B=60 \implies B=4$. Valid solution: A=7, B=4 (C=3, check: $74 \times 5 = 370$). For $A \ge 8$, $8A > 60$, no non-negative B is possible.

If $C=4$: $8A + B = 80$. If $A=1..8$, $8A+B < 80$ for $B \le 9$. If $A=9$, $72+B=80 \implies B=8$. Valid solution: A=9, B=8 (C=4, check: $98 \times 5 = 490$). For $A \ge 10$, A is not a digit.

If $C=5$: $8A + B = 100$. Maximum value of $8A+B$ for $A \in \{1..9\}$ and $B \in \{0..9\}$ is $8(9) + 9 = 72 + 9 = 81$. Since $81 < 100$, there are no solutions for $C \ge 5$.

---

The possible pairs of values for the digits A and B are:

(A, B) = (2, 4), (4, 8), (5, 0), (7, 4), (9, 8).

Test of divisibility by 2:

A number is divisible by $2$ if its units digit is an even number. The even digits are $0, 2, 4, 6,$ and $8$.

---

Now, let's check the number $90876$ for divisibility by $2$.

The given number is $90876$.

The units digit of the number $90876$ is the last digit, which is $6$.

The digit $6$ is an even number.

According to the test of divisibility by $2$, if the units digit of a number is even, the number is divisible by $2$.

Since the units digit of $90876$ is $6$, which is an even digit, the number $90876$ is divisible by $2$.

---

Justification:

Any integer can be written in the form $10k + u$, where $u$ is the units digit and $10k$ represents the rest of the number (tens, hundreds, etc.).

$10k$ is always divisible by 2 because $10k = 2 \times (5k)$.

So, the divisibility of $10k + u$ by 2 depends solely on the divisibility of $u$ by 2.

For the number $90876$, the units digit is $6$. The number can be written as $90870 + 6$.

$90870$ is divisible by $2$ (since it ends in $0$).

$6$ is divisible by $2$.

The sum of two numbers divisible by 2 is also divisible by 2.

Therefore, $90876 = 90870 + 6$ is divisible by $2$.

Test of divisibility by 3:

A number is divisible by $3$ if the sum of its digits is divisible by $3$.

---

Now, let's check the number $456789$ for divisibility by $3$.

The given number is $456789$.

Let's find the sum of its digits:

Sum of digits = $4 + 5 + 6 + 7 + 8 + 9$

Sum of digits = $9 + 6 + 7 + 8 + 9$

Sum of digits = $15 + 7 + 8 + 9$

Sum of digits = $22 + 8 + 9$

Sum of digits = $30 + 9$

Sum of digits = $39$

Now, we check if the sum of the digits, $39$, is divisible by $3$.

$39 \div 3 = 13$

Since the sum of the digits, $39$, is divisible by $3$, the number $456789$ is divisible by $3$.

---

Justification for the test of divisibility by 3:

Consider a 3-digit number represented in general form as $100a + 10b + c$, where $a, b,$ and $c$ are the digits (hundreds, tens, and units digit, respectively).

We can rewrite this number using the fact that powers of 10 can be expressed in terms of multiples of 9 (or 3) plus 1:

$10 = 9 + 1$

$100 = 99 + 1$

$1000 = 999 + 1$, and so on.

So, the number $100a + 10b + c$ can be written as:

$(99 + 1)a + (9 + 1)b + c$

Distribute the terms:

$99a + a + 9b + b + c$

Group the terms that are multiples of 9 (and thus multiples of 3) and the remaining terms:

$(99a + 9b) + (a + b + c)$

Factor out 9 from the first group:

$9(11a + b) + (a + b + c)$

The term $9(11a + b)$ is a multiple of 9, which means it is also a multiple of 3.

The second term is the sum of the digits, $(a + b + c)$.

The original number can be expressed as the sum of a number divisible by 3 (or 9) and the sum of its digits.

If the sum of the digits $(a + b + c)$ is divisible by 3, then the entire number $[9(11a + b) + (a + b + c)]$ will be the sum of two numbers divisible by 3, and therefore, the entire number will be divisible by 3.

If the sum of the digits $(a + b + c)$ is not divisible by 3, then the entire number will be the sum of a number divisible by 3 and a number not divisible by 3, which means the entire number will not be divisible by 3.

This logic extends to numbers with any number of digits.

For $456789$, the sum of digits is $39$. Since $39$ is divisible by $3$, the number $456789$ is divisible by $3$.

Test of divisibility by 5:

A number is divisible by $5$ if its units digit is either $0$ or $5$.

---

Now, let's check the number $34565$ for divisibility by $5$.

The given number is $34565$.

The units digit of the number $34565$ is the last digit, which is $5$.

According to the test of divisibility by $5$, if the units digit of a number is $0$ or $5$, the number is divisible by $5$.

Since the units digit of $34565$ is $5$, the number $34565$ is divisible by $5$.

---

Next, let's check the number $34560$ for divisibility by $5$.

The given number is $34560$.

The units digit of the number $34560$ is the last digit, which is $0$.

According to the test of divisibility by $5$, if the units digit of a number is $0$ or $5$, the number is divisible by $5$.

Since the units digit of $34560$ is $0$, the number $34560$ is divisible by $5$.

Test of divisibility by 10:

A number is divisible by $10$ if its units digit is $0$.

---

Now, let's check the number $12345$ for divisibility by $10$.

The given number is $12345$.

The units digit of the number $12345$ is the last digit, which is $5$.

According to the test of divisibility by $10$, a number is divisible by $10$ only if its units digit is $0$.

Since the units digit of $12345$ is $5$, and not $0$, the number $12345$ is not divisible by $10$.

---

Next, let's check the number $12340$ for divisibility by $10$.

The given number is $12340$.

The units digit of the number $12340$ is the last digit, which is $0$.

According to the test of divisibility by $10$, a number is divisible by $10$ if its units digit is $0$.

Since the units digit of $12340$ is $0$, the number $12340$ is divisible by $10$.

Test of divisibility by 9:

A number is divisible by $9$ if the sum of its digits is divisible by $9$.

---

Now, let's check the number $25734$ for divisibility by $9$.

The given number is $25734$.

Let's find the sum of its digits:

Sum of digits = $2 + 5 + 7 + 3 + 4$

Sum of digits = $7 + 7 + 3 + 4$

Sum of digits = $14 + 3 + 4$

Sum of digits = $17 + 4$

Sum of digits = $21$

Now, we check if the sum of the digits, $21$, is divisible by $9$.

$21 \div 9$ is not an integer. $9 \times 2 = 18$ and $9 \times 3 = 27$. $21$ lies between $18$ and $27$.

Since the sum of the digits, $21$, is not divisible by $9$, the number $25734$ is not divisible by $9$.

---

Justification for the test of divisibility by 9:

Consider a number, for example, a 3-digit number $abc = 100a + 10b + c$.

We can rewrite this as:

$100a + 10b + c = (99 + 1)a + (9 + 1)b + c$

$= 99a + a + 9b + b + c$

$= (99a + 9b) + (a + b + c)$

$= 9(11a + b) + (a + b + c)$

The term $9(11a + b)$ is always divisible by 9.

Therefore, the divisibility of the original number $100a + 10b + c$ by 9 depends entirely on the divisibility of the remaining term, which is the sum of the digits $(a + b + c)$, by 9.

If the sum of digits is divisible by 9, the number is divisible by 9. If the sum of digits is not divisible by 9, the number is not divisible by 9.

This principle applies to numbers with any number of digits.

For $25734$, the sum of digits is $21$. Since $21$ is not divisible by $9$, the number $25734$ is not divisible by $9$.

Test of divisibility by 6:

A number is divisible by $6$ if and only if it is divisible by both $2$ and $3$.

To be divisible by 2, the units digit must be an even number ($0, 2, 4, 6,$ or $8$).

To be divisible by 3, the sum of the digits must be divisible by $3$.

---

Now, let's check the number $45312$ for divisibility by $6$.

First, check divisibility by $2$. The units digit of $45312$ is $2$. Since $2$ is an even number, $45312$ is divisible by $2$.

Next, check divisibility by $3$. Calculate the sum of the digits of $45312$:

Sum of digits = $4 + 5 + 3 + 1 + 2$

Sum of digits = $9 + 3 + 1 + 2$

Sum of digits = $12 + 1 + 2$

Sum of digits = $13 + 2$

Sum of digits = $15$

Check if the sum of the digits, $15$, is divisible by $3$. $15 \div 3 = 5$. Since $15$ is divisible by $3$, the number $45312$ is divisible by $3$.

Since $45312$ is divisible by both $2$ and $3$, it is divisible by $6$.

---

Justification:

The number $6$ can be factored into two prime numbers, $2$ and $3$. These prime factors are coprime (they share no common factors other than 1).

If a number is divisible by two coprime numbers, it is also divisible by their product.

In this case, since $45312$ is divisible by $2$ (as its units digit is even) and divisible by $3$ (as the sum of its digits is divisible by $3$), and $2$ and $3$ are coprime, $45312$ must be divisible by the product of $2$ and $3$, which is $6$.

Test of divisibility by 4:

A number is divisible by $4$ if the number formed by its last two digits (tens digit and units digit) is divisible by $4$.

---

Now, let's check the number $78916$ for divisibility by $4$.

The given number is $78916$.

The number formed by the last two digits of $78916$ is $16$.

Now, we check if the number $16$ is divisible by $4$.

$16 \div 4 = 4$

Since the number formed by the last two digits ($16$) is divisible by $4$, the number $78916$ is divisible by $4$.

---

Justification for the test of divisibility by 4:

Consider a number, say a number with three or more digits. We can write this number in general form as $1000a + 100b + 10c + d$ (for a 4-digit number, where $d$ is the units digit, $c$ the tens, $b$ the hundreds, and $a$ the thousands).

We can rewrite the number as: $(1000a + 100b) + (10c + d)$.

Notice that any multiple of 100 is divisible by 4, because $100 = 4 \times 25$.

So, $1000a = 10 \times (100a) = 10 \times (4 \times 25a) = 4 \times (250a)$. This term is divisible by 4.

Similarly, $100b = 4 \times (25b)$. This term is divisible by 4.

Therefore, the part of the number represented by the digits from the hundreds place onwards (e.g., $1000a + 100b$) is always divisible by 4.

The original number can be expressed as the sum of a number divisible by 4 (the part representing hundreds and above) and the number formed by the last two digits ($10c + d$).

The original number = (Part divisible by 4) + (Number formed by last two digits).

For the entire number to be divisible by 4, the remaining part, which is the number formed by its last two digits ($10c + d$), must be divisible by 4.

For the number $78916$, it can be written as $78900 + 16$. $78900$ is divisible by $100$, so it is divisible by $4$. The number formed by the last two digits is $16$. Since $16$ is divisible by $4$, $78916$ is divisible by $4$.

Test of divisibility by $8$:

A number is divisible by $8$ if the number formed by its last three digits is divisible by $8$. This rule applies to any integer. For a number with four or more digits, its divisibility by $8$ is determined solely by the divisibility of the number formed by its hundreds, tens, and units digits by $8$. For numbers less than $1000$, the number itself must be divisible by $8$.

---

Applying the test to $123456$:

The given number is $123456$.

We need to examine the number formed by its last three digits.

The last three digits of $123456$ are $456$.

---

Justification:

Now, we check if the number $456$ is divisible by $8$.

We can perform the division:

$456 \div 8$

$456 = 8 \times 57$

Since $456$ is exactly divisible by $8$ (it gives a quotient of $57$ with no remainder), the condition for divisibility by $8$ is met.

---

Conclusion:

Based on the test of divisibility by $8$, since the number formed by the last three digits of $123456$ (which is $456$) is divisible by $8$, the number $123456$ is also divisible by $8$.

Thus, the answer is Yes, $123456$ is divisible by $8$.

Test of divisibility by $11$:

A number is divisible by $11$ if the difference between the sum of the digits at odd places (from the right) and the sum of the digits at even places (from the right) is either $0$ or a multiple of $11$.

---

Applying the test to $1331$:

The given number is $1331$.

Let's identify the digits and their positions starting from the right:

• Digit at $1^{\text{st}}$ place (odd) from the right is $1$.
• Digit at $2^{\text{nd}}$ place (even) from the right is $3$.
• Digit at $3^{\text{rd}}$ place (odd) from the right is $3$.
• Digit at $4^{\text{th}}$ place (even) from the right is $1$.

Sum of digits at odd places (from the right) $= $ Digit at $1^{\text{st}}$ place + Digit at $3^{\text{rd}}$ place $= 1 + 3 = 4$.

Sum of digits at even places (from the right) $= $ Digit at $2^{\text{nd}}$ place + Digit at $4^{\text{th}}$ place $= 3 + 1 = 4$.

---

Justification:

Now, we find the difference between these two sums:

Difference = (Sum of digits at odd places) - (Sum of digits at even places)

Difference $= 4 - 4 = 0$.

According to the test of divisibility by $11$, if this difference is $0$ or a multiple of $11$, the number is divisible by $11$.

In this case, the difference is $0$, which satisfies the condition for divisibility by $11$.

---

Conclusion:

Since the difference between the sum of the digits at odd places and the sum of the digits at even places of $1331$ is $0$, the number $1331$ is divisible by $11$.

Thus, the answer is Yes, $1331$ is divisible by $11$.

To determine if $23456$ is divisible by $4$ and $8$ without actual division, we use the tests of divisibility for $4$ and $8$.

---

Divisibility by $4$:

The test of divisibility by $4$ states that a number is divisible by $4$ if the number formed by its last two digits is divisible by $4$.

For the number $23456$, the last two digits form the number $56$.

We check if $56$ is divisible by $4$.

$56 \div 4 = 14$ with a remainder of $0$.

Since $56$ is exactly divisible by $4$, the number $23456$ is divisible by $4$.

Thus, $23456$ is divisible by $4$.

---

Divisibility by $8$:

The test of divisibility by $8$ states that a number is divisible by $8$ if the number formed by its last three digits is divisible by $8$.

For the number $23456$, the last three digits form the number $456$.

We check if $456$ is divisible by $8$.

$456 \div 8 = 57$ with a remainder of $0$.

Since $456$ is exactly divisible by $8$, the number $23456$ is divisible by $8$.

Thus, $23456$ is divisible by $8$.

Given:

The number is $21y5$, where $y$ is a digit.

The number $21y5$ is a multiple of $9$.

---

To Find:

The value of the digit $y$.

---

Solution:

We use the test of divisibility by $9$.

A number is divisible by $9$ if the sum of its digits is divisible by $9$.

The digits of the number $21y5$ are $2$, $1$, $y$, and $5$.

The sum of the digits is $2 + 1 + y + 5 = 8 + y$.

Since the number $21y5$ is a multiple of $9$, the sum of its digits must be a multiple of $9$.

So, $8 + y$ must be a multiple of $9$.

The digit $y$ can be any integer from $0$ to $9$ (i.e., $y \in \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$).

We need to find a value of $y$ such that $8+y$ is a multiple of $9$ and $y$ is a single digit.

Possible multiples of $9$ are $0, 9, 18, 27, ...$

Let's consider the possibilities for $8+y$:

If $8 + y = 0$, then $y = -8$, which is not a valid digit.

If $8 + y = 9$, then $y = 9 - 8 = 1$. This is a valid digit ($0 \leq 1 \leq 9$).

If $8 + y = 18$, then $y = 18 - 8 = 10$, which is not a valid single digit.

Any larger multiple of $9$ will result in a value of $y$ greater than $9$.

Therefore, the only valid value for $y$ is $1$.

---

Justification:

If $y=1$, the number is $2115$.

Sum of digits $= 2 + 1 + 1 + 5 = 9$.

Since $9$ is a multiple of $9$, the number $2115$ is divisible by $9$. This confirms our value of $y$ is correct.

---

Answer:

The value of $y$ is $1$.

Given:

The number is $31z5$, where $z$ is a digit.

The number $31z5$ is a multiple of $3$.

---

To Find:

The possible values of the digit $z$.

---

Solution:

We use the test of divisibility by $3$.

A number is divisible by $3$ if the sum of its digits is divisible by $3$.

The digits of the number $31z5$ are $3$, $1$, $z$, and $5$.

The sum of the digits is $3 + 1 + z + 5 = 9 + z$.

Since the number $31z5$ is a multiple of $3$, the sum of its digits must be a multiple of $3$.

So, $9 + z$ must be a multiple of $3$.

The digit $z$ can be any integer from $0$ to $9$ (i.e., $z \in \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$).

We need to find values of $z$ such that $9+z$ is a multiple of $3$ and $z$ is a single digit.

Since $9$ is already a multiple of $3$, the sum $9+z$ will be a multiple of $3$ if and only if $z$ is a multiple of $3$.

The multiples of $3$ that are single digits are $0, 3, 6, 9$.

Let's list the possible values for $z$ that make $9+z$ a multiple of $3$:

• If $z = 0$, $9 + 0 = 9$, which is a multiple of $3$. ($9 \div 3 = 3$)
• If $z = 1$, $9 + 1 = 10$, which is not a multiple of $3$.
• If $z = 2$, $9 + 2 = 11$, which is not a multiple of $3$.
• If $z = 3$, $9 + 3 = 12$, which is a multiple of $3$. ($12 \div 3 = 4$)
• If $z = 4$, $9 + 4 = 13$, which is not a multiple of $3$.
• If $z = 5$, $9 + 5 = 14$, which is not a multiple of $3$.
• If $z = 6$, $9 + 6 = 15$, which is a multiple of $3$. ($15 \div 3 = 5$)
• If $z = 7$, $9 + 7 = 16$, which is not a multiple of $3$.
• If $z = 8$, $9 + 8 = 17$, which is not a multiple of $3$.
• If $z = 9$, $9 + 9 = 18$, which is a multiple of $3$. ($18 \div 3 = 6$)

The possible single-digit values for $z$ are $0, 3, 6, 9$.

---

Answer:

The possible values of $z$ are $0, 3, 6, 9$.

Given:

The number is $1a2b$, where $a$ and $b$ are digits.

The number $1a2b$ is divisible by both $5$ and $10$.

---

To Find:

The possible values of the digit $b$.

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Solution:

We are given that the number $1a2b$ is divisible by $5$ and $10$.

First, let's consider the test of divisibility by $10$.

A number is divisible by $10$ if its last digit is $0$.

In the number $1a2b$, the last digit is $b$.

For $1a2b$ to be divisible by $10$, the digit $b$ must be $0$.

---

Next, let's consider the test of divisibility by $5$.

A number is divisible by $5$ if its last digit is either $0$ or $5$.

In the number $1a2b$, the last digit is $b$.

For $1a2b$ to be divisible by $5$, the digit $b$ must be either $0$ or $5$.

---

We are given that the number $1a2b$ must be divisible by both $5$ and $10$.

For the number to be divisible by $10$, $b$ must be $0$.

For the number to be divisible by $5$, $b$ must be $0$ or $5$.

To satisfy both conditions simultaneously, the value of $b$ must be the one that appears in both possibilities.

The common value is $0$.

Alternatively, we know that any number divisible by $10$ is also divisible by $5$ (since $10 = 2 \times 5$). Therefore, the condition of divisibility by $10$ is stronger and implies divisibility by $5$. We only need to satisfy the divisibility by $10$ test.

The divisibility test by $10$ requires the last digit to be $0$.

Therefore, $b$ must be $0$.

---

Answer:

For the number $1a2b$ to be divisible by both $5$ and $10$, the digit $b$ must be $0$.

The only possible value of $b$ is $0$.

Given:

A 2-digit number. Its units digit is $8$.

The number formed by reversing its digits is $36$ less than the original number.

---

To Find:

The original number.

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Solution:

Let the tens digit of the original 2-digit number be $t$ and the units digit be $u$.

The general form of a 2-digit number is $10 \times (\text{tens digit}) + (\text{units digit})$.

So, the original number can be written as $10t + u$.

We are given that the units digit is $8$.

Substituting the value of $u$ into the expression for the original number, we get:

Original number $= 10t + 8$

For a standard 2-digit number, the tens digit $t$ must be an integer from $1$ to $9$ ($t \in \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$).

The number formed by reversing the digits has $u$ as the tens digit and $t$ as the units digit.

The reversed number is $10u + t$.

Substituting $u=8$, the reversed number is $10(8) + t = 80 + t$.

We are given that the reversed number is $36$ less than the original number.

This means: Reversed Number = Original Number - $36$.

We can write this as an equation:

Now, we need to solve equation (i) for $t$:

$80 + t = 10t + 8 - 36$

$80 + t = 10t - 28$

Subtract $t$ from both sides of the equation:

$80 = 10t - t - 28$

$80 = 9t - 28$

Add $28$ to both sides of the equation:

$80 + 28 = 9t$

$108 = 9t$

Divide both sides by $9$ to find the value of $t$:

The calculated value for the tens digit is $t = 12$. However, for a number to be a 2-digit number, its tens digit must be a single digit from $1$ to $9$. The value $12$ is not a single digit in this range.

If we strictly follow the algebraic result obtained from the conditions given in the problem statement, we use $t=12$ to find the original number.

Original number $= 10t + 8 = 10(12) + 8 = 120 + 8 = 128$.

Let's verify this result against the problem conditions:

Original number $= 128$. Its units digit is $8$. (Condition 1 met, assuming the initial "2-digit" premise might lead to a contradiction).

Reversed number: Tens digit is $8$, units digit is $12$. If we interpret reversing digits of $128$ as switching the last digit $8$ with the "tens digit" which came out as $12$, this concept becomes ambiguous for a number like $128$. If we strictly consider the original number $10t+u$ with $u=8$, $t=12$, the number is $128$. The reversed number is formed by swapping the digits, which for 128 doesn't naturally mean swapping 8 and 12. For a number like $XY$, reversed is $YX$. For $128$, is it $821$? Or $812$? The problem likely assumes the standard 2-digit number representation $10t+u$ where $t$ and $u$ are single digits.

Given the derivation $t=12$ from the equation $80+t = 10t+8-36$, which came directly from the problem setup ($u=8$, Reversed = Original - 36), the value $t=12$ is mathematically required by the stated conditions, even though it contradicts the definition of a standard 2-digit number's tens digit.

Assuming the intent was to follow the algebraic relationship derived, the original number is calculated using $t=12$.

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Answer:

Following the conditions that the units digit is $8$ and the reversed number is $36$ less than the original number, the calculation yields a value of $12$ for the tens digit. Although this contradicts the definition of a standard 2-digit number, using this value, the original number is $10(12) + 8 = 128$. The reversed number formed by digits 8 and 12 is not standard. If we consider the number 128, its reverse in the sense of swapping first and last digits doesn't fit the calculation. However, the calculation $t=12$ is directly implied by the given relations for the 'tens digit' $t$ and 'units digit' $u=8$ in the structure $10t+u$.

Based on the mathematical derivation from the problem statement, the original number is calculated as $128$.

Given:

Addition: $\begin{array}{cc} & 1 & A \\ + & B & 8 \\ \hline & 6 & C \\ \hline \end{array}$

No carry-over from the units place.

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Solution:

Consider the units column:

Since there is no carry-over, $A + 8 < 10$. As A is a digit ($0 \leq A \leq 9$), this implies $A < 2$. So, A can be $0$ or $1$.

Consider the tens column:

The carry from the units place is $0$.

Solving for B, we get $B = 6 - 1 = 5$.

Now use the possible values for A and the value of B to find C.

From the units column, $A + 8 = C$.

If $A=0$, then $C = 0 + 8 = 8$. This gives the solution A=0, B=5, C=8.

If $A=1$, then $C = 1 + 8 = 9$. This gives the solution A=1, B=5, C=9.

Both sets of values (A=0, B=5, C=8) and (A=1, B=5, C=9) satisfy the conditions. Assuming the common practice where the letters should form a valid addition puzzle, the first solution (A=0, B=5, C=8) is typically presented.

---

Answer:

The values are:

$A = 0$

$B = 5$

$C = 8$

Given:

The multiplication problem:

$\begin{array}{c} \phantom{\times} A \\ \times \phantom{0} 4 \\ \hline 1 A \\ \hline \end{array}$

Here, A is a single digit.

---

To Find:

The value of the digit A.

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Solution:

In this multiplication puzzle, A represents a single digit (an integer from $0$ to $9$). The top line represents the number A. The result of the multiplication $A \times 4$ is a 2-digit number represented by '1A'.

According to standard notation for such puzzles, the number '1A' represents a two-digit number where the tens digit is $1$ and the units digit is $A$. The value of this number is $1 \times 10 + A$.

So, we can write the given multiplication problem as an algebraic equation:

Now, we need to solve this equation for A, keeping in mind that A must be a single digit ($0 \leq A \leq 9$).

Subtract A from both sides of equation (i):

$4A - A = 10$

$3A = 10$

Now, divide both sides by $3$ to find the value of A:

The value we found for A is $\frac{10}{3}$, which is $3.\overline{3}$. This is not an integer, and therefore, it cannot be a single digit from $0$ to $9$.

Let's also consider the problem using column multiplication with carry-over.

Units column: $4 \times A$ results in a units digit of A and some carry-over $c_u$ to the tens place.

So, $4 \times A = 10 \times c_u + A$.

This simplifies to $3A = 10 c_u$.

Since A is a single digit ($0 \leq A \leq 9$), the possible values for $3A$ are $\{0, 3, 6, 9, 12, 15, 18, 21, 24, 27\}$. For $3A$ to be equal to $10 \times c_u$ (where $c_u$ is an integer carry), $3A$ must be a multiple of $10$. The only multiple of $10$ in the list of possible values for $3A$ is $0$.

If $3A = 0$, then $A = 0$. In this case, $10 c_u = 0$, so $c_u = 0$.

Thus, from the units column, we get $A=0$ and a carry-over $c_u=0$.

Tens column: The top number A is a single digit, meaning its tens digit is $0$. The multiplication in the tens column involves $4 \times (\text{tens digit of A}) + (\text{carry-over from units})$. The result should be the tens digit of the final answer, which is $1$.

$4 \times 0 + c_u = 1$

$0 + c_u = 1$

$c_u = 1$.

From the units column calculation, we found the carry-over $c_u$ must be $0$. From the tens column position in the result, we found the carry-over $c_u$ must be $1$. These two findings ($c_u = 0$ and $c_u = 1$) contradict each other.

Both the direct algebraic translation and the column multiplication analysis lead to the conclusion that there is no single digit A that satisfies the conditions of the given multiplication problem.

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Conclusion:

Based on the standard interpretation of such mathematical puzzles, there is no integer digit A (from $0$ to $9$) that solves the given multiplication problem. The equation $4A = 10+A$ derived from the problem layout does not yield a single-digit integer solution for A.

Let the 3-digit number be represented by its digits p, q, and r, where p is the hundreds digit, q is the tens digit, and r is the units digit.

For a number to be a 3-digit number, the hundreds digit cannot be zero. So, $p \in \{1, 2, ..., 9\}$, and $q, r \in \{0, 1, ..., 9\}$.

---

General Form of the number 'pqr':

The number 'pqr' can be written in its general form by considering the place value of each digit:

• p is in the hundreds place, its value is $p \times 100$.
• q is in the tens place, its value is $q \times 10$.
• r is in the units place, its value is $r \times 1$.

The general form of the number 'pqr' is the sum of the values of its digits:

$100p + 10q + r$

---

General Form of the number 'rqp' (reversed digits):

The number formed by reversing the digits of 'pqr' is 'rqp'. In this new number:

• r is the hundreds digit.
• q is the tens digit.
• p is the units digit.

For 'rqp' to be a 3-digit number, the hundreds digit r must not be zero. So, $r \in \{1, 2, ..., 9\}$, and $q, p \in \{0, 1, ..., 9\}$. Note that if the original number was 'pqr' with $p \neq 0$, the reversed number 'rqp' is a 3-digit number only if $r \neq 0$.

The general form of the number 'rqp' is:

$100r + 10q + p$

---

Summary:

General form of 'pqr': $100p + 10q + r$

General form of 'rqp': $100r + 10q + p$

Given:

A 3-digit number represented as 'abc', where a, b, and c are digits. 'abc' represents the number with a in the hundreds place, b in the tens place, and c in the units place. It is given that it is a 3-digit number, so $a \neq 0$. The numbers 'bca' and 'cab' are formed by cyclically permuting the digits.

---

To Prove:

The sum of the numbers 'abc', 'bca', and 'cab' is always divisible by 111.

That is, $('abc') + ('bca') + ('cab')$ is divisible by $111$.

---

Proof:

Let the 3-digit number be 'abc'. Using the general form, this number can be written as:

The number formed by cyclically shifting the digits once to the left is 'bca'. Its general form is:

The number formed by cyclically shifting the digits again (or shifting from 'abc' twice to the left) is 'cab'. Its general form is:

Now, let's find the sum of these three numbers:

Sum $= \text{'abc'} + \text{'bca'} + \text{'cab'}$

Sum $= (100a + 10b + c) + (100b + 10c + a) + (100c + 10a + b)$

Group the terms containing the same digits (a, b, and c):

Sum $= (100a + a + 10a) + (10b + 100b + b) + (c + 10c + 100c)$

Combine the coefficients of a, b, and c:

Sum $= (100 + 1 + 10)a + (10 + 100 + 1)b + (1 + 10 + 100)c$

Sum $= 111a + 111b + 111c$

Factor out the common factor $111$ from the sum:

Since the sum can be expressed as $111$ multiplied by an integer $(a+b+c)$, the sum is always divisible by $111$. This holds true for any digits a, b, and c, where a is not zero.

---

Example:

Let's take the 3-digit number $123$. Here, $a=1$, $b=2$, and $c=3$.

The number 'abc' is $123$.

The number 'bca' is $231$.

The number 'cab' is $312$.

Now, let's find their sum:

Sum $= 123 + 231 + 312 = 666$.

Using the proven property, the sum should be $111(a+b+c)$.

Here, $a+b+c = 1+2+3 = 6$.

So, $111(a+b+c) = 111 \times 6 = 666$.

Now, let's check if the sum $666$ is divisible by $111$:

$666 \div 111 = 6$

Since the division results in an integer (6) with no remainder, the sum $666$ is divisible by $111$.

This example illustrates the proven property.

Given:

A 3-digit number 'abc', where a, b, and c are digits. 'abc' represents the number with a in the hundreds place, b in the tens place, and c in the units place. It is given that it is a 3-digit number, so $a \neq 0$. The number 'cba' is formed by reversing the digits, which means c is the hundreds digit, b is the tens digit, and a is the units digit. For 'cba' to be a 3-digit number, $c \neq 0$. The problem states the assumption $a > c$. This also implies $a \neq 0$ and $c \neq 0$.

---

To Prove:

The difference between 'abc' and 'cba' is always divisible by $99$ and $9$, assuming $a > c$.

That is, $('abc') - ('cba')$ is divisible by $99$ and $9$.

---

Proof:

Let the 3-digit number be 'abc'. Using the general form, this number can be written as:

The number formed by reversing the digits is 'cba'. Its general form is:

Now, let's find the difference between these two numbers, assuming $a > c$ as given, so 'abc' is the larger number:

Difference $= \text{'abc'} - \text{'cba'}$

Difference $= (100a + 10b + c) - (100c + 10b + a)$

Remove the parentheses and group the terms containing the same digits (a, b, and c):

Difference $= 100a + 10b + c - 100c - 10b - a$

Difference $= (100a - a) + (10b - 10b) + (c - 100c)$

Combine the coefficients of a, b, and c:

Difference $= 99a + 0b - 99c$

Factor out the common factor $99$ from the difference:

Since the difference can be expressed as $99$ multiplied by an integer $(a - c)$, the difference is always divisible by $99$.

Also, since $99 = 9 \times 11$, we can write:

Since the difference can be expressed as $9$ multiplied by an integer ($11 \times (a - c)$), the difference is always divisible by $9$.

This holds true for any digits a and c, where a is not zero, c is not zero (for 'cba' to be 3-digit), and $a > c$.

---

Example:

Let's take the 3-digit number $573$. Here, $a=5$, $b=7$, and $c=3$. We have $a > c$ ($5 > 3$).

The number 'abc' is $573$.

The number 'cba' is $375$.

Now, let's find their difference:

Difference $= 573 - 375$.

$\begin{array}{cc} & 5 & 7 & 3 \\ - & 3 & 7 & 5 \\ \hline & 1 & 9 & 8 \\ \hline \end{array}$

Difference $= 198$.

Using the proven property, the difference should be $99(a-c)$.

Here, $a-c = 5-3 = 2$.

So, $99(a-c) = 99 \times 2 = 198$. This matches our calculation.

Now, let's check if the difference $198$ is divisible by $99$ and $9$:

Divisibility by $99$: $198 \div 99 = 2$. Since the division results in an integer, $198$ is divisible by $99$.

Divisibility by $9$: $198 \div 9 = 22$. Since the division results in an integer, $198$ is divisible by $9$.

This example illustrates the proven property.

Given:

The addition problem:

$\begin{array}{cc} & A & B \\ + & 8 & 7 \\ \hline & C & A & 3 \\ \hline \end{array}$

Here, A, B, and C represent single digits from $0$ to $9$. In the result 'CA3', C is the hundreds digit, A is the tens digit, and 3 is the units digit.

---

To Find:

The values of the digits A, B, and C.

---

Solution:

We will solve this cryptarithmetic puzzle by analysing the columns from right to left (units to hundreds).

Step 1: Analyse the Units Column

The digits in the units column are B and 7. Their sum results in a units digit of 3 in the total sum. Let the carry-over to the tens column be $c_1$.

Since B is a single digit ($0 \leq B \leq 9$), the possible values for $B+7$ are integers from $0+7=7$ to $9+7=16$.

We need $B+7$ to be a number whose units digit is 3. The only number in the range $[7, 16]$ with a units digit of 3 is 13.

So, $B + 7 = 13$.

Solving for B:

This is a valid digit ($0 \leq 6 \leq 9$). From equation (i), if $B+7=13$, then $13 = 3 + 10 \times c_1$, which means $10 \times c_1 = 10$, so $c_1 = 1$.

Thus, from the units column, we get $B=6$ and a carry-over of $c_1=1$ to the tens column.

Step 2: Analyse the Tens Column

The digits in the tens column are A and 8, plus the carry-over $c_1$ from the units column. Their sum results in the tens digit of the total sum, which is given as A in the result 'CA3'. Let the carry-over from the tens column to the hundreds column be $c_2$.

Here, the 'A' on the left is the tens digit of the first number, and the 'A' on the right is the value of the tens digit of the sum 'CA3'. Substitute the value $c_1=1$ into equation (ii):

Subtract A from both sides of the equation:

Solving for $c_2$:

The carry-over $c_2$ must be an integer (0 or 1 in this case, as the sum is a 3-digit number). However, we calculated $c_2 = 0.9$, which is not an integer.

Let's re-examine the interpretation of the tens column sum resulting in a tens digit A. The sum of the tens column digits ($A + 8 + c_1$) gives a number whose units digit must be A, and whose tens digit is the carry-over $c_2$.

Sum of tens column digits $= A + 8 + 1 = A + 9$.

Let $S_T = A + 9$. We require the units digit of $S_T$ to be A, and the tens digit of $S_T$ to be $c_2$.

We test the possible single-digit values for A ($0 \leq A \leq 9$):

• If A = 0, $S_T = 0 + 9 = 9$. Units digit is 9. We need units digit to be A=0. $9 \neq 0$. No.
• If A = 1, $S_T = 1 + 9 = 10$. Units digit is 0. We need units digit to be A=1. $0 \neq 1$. No.
• If A = 2, $S_T = 2 + 9 = 11$. Units digit is 1. We need units digit to be A=2. $1 \neq 2$. No.
• If A = 3, $S_T = 3 + 9 = 12$. Units digit is 2. We need units digit to be A=3. $2 \neq 3$. No.
• If A = 4, $S_T = 4 + 9 = 13$. Units digit is 3. We need units digit to be A=4. $3 \neq 4$. No.
• If A = 5, $S_T = 5 + 9 = 14$. Units digit is 4. We need units digit to be A=5. $4 \neq 5$. No.
• If A = 6, $S_T = 6 + 9 = 15$. Units digit is 5. We need units digit to be A=6. $5 \neq 6$. No.
• If A = 7, $S_T = 7 + 9 = 16$. Units digit is 6. We need units digit to be A=7. $6 \neq 7$. No.
• If A = 8, $S_T = 8 + 9 = 17$. Units digit is 7. We need units digit to be A=8. $7 \neq 8$. No.
• If A = 9, $S_T = 9 + 9 = 18$. Units digit is 8. We need units digit to be A=9. $8 \neq 9$. No.

In every case, the units digit of $(A+9)$ is not equal to A. This means there is no single digit A that satisfies the condition for the tens column based on the standard rules of column addition.

Step 3: Analyse the Hundreds Column

The hundreds digit of the result is C. This digit comes from the carry-over $c_2$ from the tens column.

Since our analysis of the tens column shows no valid integer value for $c_2$, we cannot determine a value for C either.

---

Conclusion:

Our step-by-step analysis of the column addition, following standard rules of arithmetic and the given puzzle structure where A, B, and C are single digits, leads to a contradiction in the tens column. Specifically, the requirement that the sum of the tens column ($A+8+1$) must have a units digit equal to A is not satisfied by any single digit value of A.

Therefore, based on the standard interpretation of this type of puzzle, there are no single digit values for A, B, and C that solve the given addition problem.

Given:

The multiplication problem:

$\begin{array}{cc} & A & B \\ \times & & 3 \\ \hline C & A & B \\ \hline \end{array}$

Here, A, B, and C represent single digits from $0$ to $9$. The number 'AB' represents $10A + B$, and the number 'CAB' represents $100C + 10A + B$. Since 'AB' is at the top, A cannot be $0$ for it to be typically written this way as a 2-digit number. Similarly, since 'CAB' is a 3-digit number, C cannot be $0$. Thus, $A \in \{1, 2, ..., 9\}$, $B \in \{0, 1, ..., 9\}$, and $C \in \{1, 2, ..., 9\}$.

---

To Find:

The values of the digits A, B, and C.

---

Solution:

We solve this puzzle by considering the multiplication column by column, starting from the units place and moving to the left, accounting for any carry-overs.

Step 1: Analyse the Units Column

We multiply the units digit of 'AB' (which is B) by 3. The units digit of the result must be the units digit of 'CAB' (which is B). Let the carry-over to the tens column be $c_1$.

The multiplication in the units column gives:

In terms of place values and carry-over, this can be written as:

Subtract B from both sides of the equation:

3B - B = $10c_1$

2B = $10c_1$

Divide both sides by 2:

Since B is a single digit ($0 \leq B \leq 9$), we can find the possible values for B by considering possible integer values for the carry-over $c_1$. Since $3 \times B \leq 3 \times 9 = 27$, the maximum possible carry $c_1$ can be 2 (from $3 \times 9 = 27 = 10 \times 2 + 7$).

• If $c_1 = 0$, B = $5 \times 0 = 0$. This is a valid digit. If $B=0$, $3 \times 0 = 0$. Units digit is 0, carry $c_1=0$. This matches $B=0$.
• If $c_1 = 1$, B = $5 \times 1 = 5$. This is a valid digit. If $B=5$, $3 \times 5 = 15$. Units digit is 5, carry $c_1=1$. This matches $B=5$.
• If $c_1 = 2$, B = $5 \times 2 = 10$. This is not a single digit. So $c_1$ cannot be 2 or higher.

So, there are two possible cases from the units column: ($B=0, c_1=0$) or ($B=5, c_1=1$).

Step 2: Analyse the Tens Column

We multiply the tens digit of 'AB' (which is A) by 3 and add the carry-over $c_1$ from the units column. The result is a number whose units digit is the tens digit of 'CAB' (which is A), and whose tens digit is the carry-over $c_2$ to the hundreds column.

The multiplication in the tens column gives:

In terms of place values and carry-over, this can be written as:

Subtract A from both sides:

Step 3: Analyse the Hundreds Column

The hundreds digit of the result 'CAB' is C. This digit comes solely from the carry-over $c_2$ from the tens column.

Since 'CAB' is a 3-digit number, C cannot be 0. Thus, the carry-over $c_2$ must be an integer from 1 to 9 ($c_2 \in \{1, 2, ..., 9\}$).

Step 4: Combine the Cases and Solve

We use the possible cases from Step 1 ($B=0, c_1=0$) or ($B=5, c_1=1$) and substitute them into the equation from Step 2 ($2A + c_1 = 10c_2$), also considering $c_2 \in \{1, 2, ..., 9\}$.

Case 1: $B=0$, $c_1=0$

Substitute $c_1=0$ into $2A + c_1 = 10c_2$:

$2A + 0 = 10c_2$

$2A = 10c_2$

We know $c_2$ must be in $\{1, 2, ..., 9\}$ (since $c_2=C$ and $C \neq 0$). Let's test values for $c_2$:

• If $c_2 = 1$, A = $5 \times 1 = 5$. This is a valid digit ($0 \leq 5 \leq 9$). Also, $A=5$ is not 0, satisfying the 2-digit condition for 'AB'. And $C = c_2 = 1$, which is not 0, satisfying the 3-digit condition for 'CAB'.
• If $c_2 = 2$, A = $5 \times 2 = 10$. This is not a single digit. Higher values of $c_2$ will result in even larger values for A, which are not single digits.

So, from Case 1, we get the solution A=5, B=0, and C=1 (since $C=c_2=1$).

Let's verify this solution:

$\begin{array}{cc} & 5 & 0 \\ \times & & 3 \\ \hline 1 & 5 & 0 \\ \hline \end{array}$

$50 \times 3 = 150$. Here, A=5, B=0, C=1. The result 'CAB' is '150', which matches. The conditions A $\neq$ 0, C $\neq$ 0 are met.

Case 2: $B=5$, $c_1=1$

Substitute $c_1=1$ into $2A + c_1 = 10c_2$:

The left side of the equation, $2A + 1$, is always an odd number for any integer value of A (since $2A$ is even, $2A+1$ is odd). The right side of the equation, $10c_2$, is always an even number for any integer value of $c_2$ (since it's a multiple of 10). An odd number cannot be equal to an even number. Therefore, there is no integer value for A (or $c_2$) that can satisfy this equation.

Thus, Case 2 does not yield a valid solution.

The only solution is derived from Case 1.

---

Answer:

Based on the step-by-step analysis of the multiplication, the unique solution is:

$A = 5$

$B = 0$

$C = 1$

Given:

A 2-digit number.

Condition 1: The number is three times the sum of its digits.

Condition 2: If 45 is added to the number, the digits are reversed.

---

To Find:

The original 2-digit number.

---

Solution:

Let the 2-digit number be represented by its tens digit $t$ and its units digit $u$.

The general form of the original number is $10t + u$.

Since it's a 2-digit number, the tens digit $t$ must be a non-zero digit, so $t \in \{1, 2, ..., 9\}$. The units digit $u$ can be any digit from $0$ to $9$, so $u \in \{0, 1, ..., 9\}$.

The sum of the digits of the original number is $t + u$.

The number formed by reversing the digits has $u$ as the tens digit and $t$ as the units digit. The general form of the reversed number is $10u + t$. For the reversed number to potentially be a 2-digit number (which it will be if the original tens digit $t$ is not 0), we need $u \neq 0$. However, the problem states "the digits are reversed", implying we form the number $10u+t$ regardless of whether $u$ is zero, which will be a 1-digit number if $u=0$. Let's proceed assuming $10u+t$ is the number formed by reversing, whether it's 1 or 2 digits.

Now, we translate the given conditions into equations using the general form.

Condition 1: The number is three times the sum of its digits.

Simplify equation (i):

$10t + u = 3t + 3u$

Subtract $3t$ and $u$ from both sides:

$10t - 3t = 3u - u$

From equation (ii), we see that $2u$ must be a multiple of 7. Since u is a single digit ($0 \leq u \leq 9$), the possible values for $2u$ are $\{0, 2, 4, 6, 8, 10, 12, 14, 16, 18\}$. The only multiple of 7 in this set is 14.

If $2u = 14$, then $u = 7$.

Substitute $2u = 14$ into $7t = 2u$:

$7t = 14$

$t = \frac{14}{7}$

$t = 2$.

This is a valid tens digit ($1 \leq 2 \leq 9$).

If $2u = 0$, then $u = 0$. Substitute into $7t = 2u$: $7t=0$, so $t=0$. This would mean the original number is 00, which is not a 2-digit number. So, this case is not possible.

Thus, from Condition 1, the only possible digits for the original number are $t=2$ and $u=7$. The original number is $10(2) + 7 = 27$.

Let's check if the number 27 satisfies Condition 1:

Number = 27.

Sum of digits = $2 + 7 = 9$.

Three times the sum of digits $= 3 \times 9 = 27$.

Since $27 = 27$, Condition 1 is satisfied.

Condition 2: If 45 is added to the number, the digits are reversed.

Original number + 45 = Reversed number.

Simplify equation (iii):

$10t + u + 45 = 10u + t$

Subtract $t$ and $u$ from both sides:

$10t - t + 45 = 10u - u$

Divide both sides of equation (iv) by 9:

$\frac{9t}{9} + \frac{45}{9} = \frac{9u}{9}$

Now we have two equations from the two conditions:

We can substitute equation (v) into equation (ii) to solve for $t$ and $u$. Substitute $u = t+5$ into $7t = 2u$:

$7t = 2(t + 5)$

$7t = 2t + 10$

Subtract $2t$ from both sides:

$7t - 2t = 10$

$5t = 10$

$t = \frac{10}{5}$

Now substitute the value of $t$ into equation (v) to find $u$:

$u = t + 5$

$u = 2 + 5$

The tens digit is $t=2$ and the units digit is $u=7$. Both are valid digits, and $t=2$ is non-zero.

The original number is $10t + u = 10(2) + 7 = 20 + 7 = 27$.

Let's verify if the number 27 satisfies Condition 2:

Original number = 27.

Add 45: $27 + 45 = 72$.

The reversed number of 27 is 72. The digits are indeed reversed.

Since the number 27 satisfies both conditions, it is the correct answer.

---

Answer:

The original 2-digit number is $27$.

Given:

A 2-digit number.

Condition 1: The sum of its digits is 12.

Condition 2: The number obtained by interchanging the digits is 54 more than the original number.

---

To Find:

The original 2-digit number.

---

Solution:

Let the tens digit of the original 2-digit number be $t$ and the units digit be $u$.

The general form of the original number is $10t + u$.

For a 2-digit number, $t \in \{1, 2, ..., 9\}$ and $u \in \{0, 1, ..., 9\}$.

The number obtained by interchanging the digits has $u$ as the tens digit and $t$ as the units digit.

The general form of the number with interchanged digits is $10u + t$.

Now, we translate the given conditions into equations using the general form.

Condition 1: The sum of the digits is 12.

Condition 2: The number obtained by interchanging the digits is 54 more than the original number.

Interchanged number = Original number + 54

Simplify equation (ii):

$10u + t = 10t + u + 54$

Subtract $t$ and $u$ from both sides to group terms with $t$ and $u$:

$10u - u = 10t - t + 54$

$9u = 9t + 54$

Divide the entire equation by 9 to simplify:

$\frac{9u}{9} = \frac{9t}{9} + \frac{54}{9}$

Now we have a system of two linear equations with two variables, $t$ and $u$:

We can use substitution to solve this system. Substitute the expression for $u$ from equation (iii) into equation (i):

$t + (t + 6) = 12$

$2t + 6 = 12$

Subtract 6 from both sides:

$2t = 12 - 6$

$2t = 6$

Divide by 2:

Now that we have the value of $t$, substitute it back into equation (i) or (iii) to find the value of $u$. Using equation (i):

$t + u = 12$

$3 + u = 12$

Subtract 3 from both sides:

$u = 12 - 3$

The tens digit is $t=3$ and the units digit is $u=9$. Both are valid digits, and $t=3$ is non-zero.

The original number is $10t + u = 10(3) + 9 = 30 + 9 = 39$.

Let's verify if the number 39 satisfies both conditions:

Check Condition 1: Sum of digits = $3 + 9 = 12$. Correct.

Check Condition 2: Original number = 39. Interchanged number = $10(9) + 3 = 93$. Is $93 = 39 + 54$? $39 + 54 = 93$. Correct.

Since the number 39 satisfies both conditions, it is the correct answer.

---

Answer:

The original 2-digit number is $39$.

Test of divisibility by $9$:

A number is divisible by $9$ if the sum of its digits is divisible by $9$. If the sum of the digits is not divisible by $9$, then the number is not divisible by $9$. The remainder obtained when a number is divided by $9$ is the same as the remainder obtained when the sum of its digits is divided by $9$.

---

Proof using the general form of a 3-digit number:

Let the 3-digit number be represented by its digits $a$, $b$, and $c$, where $a$ is the hundreds digit, $b$ is the tens digit, and $c$ is the units digit. Since it is a 3-digit number, $a \neq 0$. The digits $a, b, c$ are integers from $0$ to $9$.

The general form of this 3-digit number is:

We want to relate this number to the sum of its digits, which is $a + b + c$.

Let's rewrite the coefficients $100$ and $10$ in terms of multiples of $9$ plus $1$:

Substitute these into the general form of the number:

Distribute $a$ and $b$ into the parentheses:

Rearrange the terms to group the multiples of $9$ and the sum of digits:

Factor out $9$ from the first group:

Let the sum of the digits be $S = a + b + c$.

Let the integer $K = 11a + b$.

Then the number can be written as:

This equation shows that the 3-digit number is equal to a multiple of $9$ ($9K$) plus the sum of its digits ($S$).

If the sum of the digits, $S = a + b + c$, is divisible by $9$, it means $S$ is a multiple of $9$. We can write $S = 9M$ for some integer $M$.

Substitute this into the equation for the Number:

Factor out $9$ from the right side:

Since $K = 11a + b$ and $M$ are integers, their sum $(K + M)$ is also an integer. Therefore, the Number is $9$ times an integer, which means the Number is divisible by $9$.

Thus, we have proven that if the sum of the digits of a 3-digit number is divisible by $9$, the number itself is divisible by $9$. The same logic can be extended to numbers with any number of digits.

Test of divisibility by $3$:

A number is divisible by $3$ if the sum of its digits is divisible by $3$. If the sum of the digits is not divisible by $3$, then the number is not divisible by $3$. The remainder obtained when a number is divided by $3$ is the same as the remainder obtained when the sum of its digits is divided by $3$.

---

Proof using the general form of a 2-digit number:

Given:

A 2-digit number represented as $10a + b$, where $a$ is the tens digit and $b$ is the units digit. Since it is a 2-digit number, $a \in \{1, 2, ..., 9\}$ and $b \in \{0, 1, ..., 9\}$.

To Prove:

The number $10a + b$ is divisible by $3$ if the sum of its digits, $a + b$, is divisible by $3$.

---

Proof:

Let the 2-digit number be $10a + b$.

The sum of its digits is $a + b$.

We can rewrite the number $10a + b$ by splitting the coefficient of $a$:

Rearrange the terms to group the term divisible by $3$ and the sum of the digits:

Let $S$ be the sum of the digits, so $S = a + b$.

The number can be written as:

We know that $9a$ is always divisible by $3$, since $9a = 3 \times (3a)$.

From the expression Number $= 9a + S$, we can see that the divisibility of the Number by $3$ depends on the divisibility of the term $S = a + b$ by $3$, since $9a$ is already divisible by $3$.

If the sum of the digits, $S = a + b$, is divisible by $3$, it means $S$ can be written as $S = 3M$ for some integer $M$.

Substitute this into the expression for the Number:

Factor out $3$ from the right side:

Since $3a$ and $M$ are integers, their sum $(3a + M)$ is also an integer. Therefore, the Number is $3$ times an integer, which means the Number is divisible by $3$.

Thus, we have proven that if the sum of the digits of a 2-digit number is divisible by $3$, the number itself is divisible by $3$. This proof can be extended to numbers with more digits by using the same principle: expressing powers of 10 as a multiple of 3 (or 9) plus 1.

Test of divisibility by $11$:

A number is divisible by $11$ if the difference between the sum of the digits at odd places (starting from the rightmost digit as the $1^{\text{st}}$ place) and the sum of the digits at even places (starting from the rightmost digit as the $2^{\text{nd}}$ place) is either $0$ or a multiple of $11$.

Alternatively, you can find the alternating sum of the digits of the number, starting from the rightmost digit, and check if the result is divisible by $11$. The alternating sum means adding the first digit (from the right), subtracting the second, adding the third, subtracting the fourth, and so on.

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Applying the test to $121332$:

The given number is $121332$.

Let's list the digits from right to left and their positions:

• Digit at $1^{\text{st}}$ place (odd) is $2$.
• Digit at $2^{\text{nd}}$ place (even) is $3$.
• Digit at $3^{\text{rd}}$ place (odd) is $3$.
• Digit at $4^{\text{th}}$ place (even) is $1$.
• Digit at $5^{\text{th}}$ place (odd) is $2$.
• Digit at $6^{\text{th}}$ place (even) is $1$.

Calculation of the alternating sum of digits (starting from the right):

Alternatively, using the difference between sums of digits at odd and even places:

Sum of digits at odd places (from the right) $= 2 + 3 + 2 = 7$.

Sum of digits at even places (from the right) $= 3 + 1 + 1 = 5$.

Difference $= (\text{Sum of odd place digits}) - (\text{Sum of even place digits}) = 7 - 5 = 2$.

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Conclusion:

The result of the alternating sum (or the difference between the sums of digits at odd and even places) is $2$.

According to the test of divisibility by $11$, the number is divisible by $11$ if this result is $0$ or a multiple of $11$.

Since $2$ is neither $0$ nor a multiple of $11$, the number $121332$ is not divisible by $11$.

Thus, the answer is No, $121332$ is not divisible by $11$.

Given:

The multiplication problem:

$\begin{array}{cc} & A & B \\ \times & & 3 \\ \hline C & A & B \\ \hline \end{array}$

Here, A, B, and C represent single digits from $0$ to $9$. The number 'AB' represents $10A + B$. Since it's the top number in a multiplication resulting in a 3-digit number, A is likely non-zero. The number 'CAB' represents $100C + 10A + B$. Since 'CAB' is a 3-digit number, C must be non-zero. Thus, $A \in \{1, 2, ..., 9\}$, $B \in \{0, 1, ..., 9\}$, and $C \in \{1, 2, ..., 9\}$.

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To Find:

The values of the digits A, B, and C.

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Solution:

We solve this cryptarithmetic puzzle by analysing the multiplication process column by column, starting from the units place.

Step 1: Analyse the Units Column

We multiply the units digit of 'AB' (which is B) by 3. The units digit of the result of this multiplication must be the units digit of 'CAB' (which is also B). Let the carry-over to the tens column be $c_1$.

The multiplication in the units column is $3 \times B$. The units digit of this product is B.

Subtract B from both sides of the equation:

$3B - B = 10c_1$

$2B = 10c_1$

Divide both sides by 2:

Since B is a single digit ($0 \leq B \leq 9$), we check possible integer values for the carry-over $c_1$. The maximum possible product in the units column is $3 \times 9 = 27$, so the carry $c_1$ can be at most 2.

• If $c_1 = 0$, then $B = 5 \times 0 = 0$. This is a valid digit. (Check: $3 \times 0 = 0$. Units digit is 0, carry is 0).
• If $c_1 = 1$, then $B = 5 \times 1 = 5$. This is a valid digit. (Check: $3 \times 5 = 15$. Units digit is 5, carry is 1).
• If $c_1 = 2$, then $B = 5 \times 2 = 10$. This is not a single digit.

So, we have two possible cases for the units digit B and the carry-over $c_1$: ($B=0, c_1=0$) or ($B=5, c_1=1$).

Step 2: Analyse the Tens Column

We multiply the tens digit of 'AB' (which is A) by 3 and add the carry-over $c_1$ from the units column. The units digit of this result must be the tens digit of 'CAB' (which is A). The tens digit of this result is the carry-over $c_2$ to the hundreds column.

Subtract A from both sides of the equation:

$3A - A + c_1 = 10c_2$

Step 3: Analyse the Hundreds Column

The hundreds digit of the result 'CAB' is C. This digit comes solely from the carry-over $c_2$ from the tens column.

Since 'CAB' is a 3-digit number, C must be a non-zero digit ($C \in \{1, 2, ..., 9\}$). Therefore, the carry-over $c_2$ must be an integer from 1 to 9 ($c_2 \in \{1, 2, ..., 9\}$).

Step 4: Combine the Cases and Solve

Now we test the possible cases for $(B, c_1)$ derived in Step 1 and substitute them into the equation from Step 2 ($2A + c_1 = 10c_2$), while also using the constraint on $c_2$ from Step 3 ($c_2 \in \{1, 2, ..., 9\}$) and on A ($A \in \{1, 2, ..., 9\}$).

Case 1: From Units Column, $B=0$ and $c_1=0$.

Substitute $c_1 = 0$ into the equation $2A + c_1 = 10c_2$:

$2A + 0 = 10c_2$

$2A = 10c_2$

$A = 5c_2$

We know $c_2$ must be in the set $\{1, 2, ..., 9\}$. Let's find the corresponding values for A:

• If $c_2 = 1$, then $A = 5 \times 1 = 5$. This is a valid digit ($A=5 \in \{1, 2, ..., 9\}$). Also, $C = c_2 = 1$, which is a valid non-zero digit ($C=1 \in \{1, 2, ..., 9\}$). This gives the potential solution A=5, B=0, C=1.
• If $c_2 = 2$, then $A = 5 \times 2 = 10$. This is not a single digit. Any higher value for $c_2$ will also result in a value for A that is not a single digit.

So, from Case 1, the only possible solution is A=5, B=0, C=1.

Case 2: From Units Column, $B=5$ and $c_1=1$.

Substitute $c_1 = 1$ into the equation $2A + c_1 = 10c_2$:

$2A + 1 = 10c_2$

The left side of this equation, $2A + 1$, is always an odd integer for any integer value of A (since $2A$ is even, $2A+1$ is odd). The right side of the equation, $10c_2$, is always an even integer for any integer value of $c_2$ (since it is a multiple of 10). An odd number cannot be equal to an even number.

Therefore, there is no integer value for A (or $c_2$) that can satisfy the equation $2A + 1 = 10c_2$. This case yields no valid solution.

Step 5: Verify the Solution

The only valid solution found is A=5, B=0, C=1.

Let's substitute these values into the original multiplication problem:

The number 'AB' is $10A + B = 10(5) + 0 = 50$.

The multiplication is $50 \times 3$.

$50 \times 3 = 150$.

The result is 150. The structure 'CAB' means the result should have C as the hundreds digit, A as the tens digit, and B as the units digit.

In the result 150, the hundreds digit is 1, the tens digit is 5, and the units digit is 0.

This matches our values: C=1, A=5, B=0.

The constraints ($A \neq 0$, $C \neq 0$) are also satisfied since $A=5$ and $C=1$.

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Answer:

The values of the digits are:

$A = 5$

$B = 0$

$C = 1$

Given:

A 3-digit number with the following properties:

1. The sum of its digits is 18.

2. The number formed by interchanging the hundreds and units digits is 594 less than the original number.

3. The units digit is twice the hundreds digit.

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To Find:

The original 3-digit number.

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Solution:

Let the hundreds digit of the 3-digit number be $h$, the tens digit be $t$, and the units digit be $u$.

The 3-digit number can be expressed in the general form as $100h + 10t + u$.

According to the problem statement, we can set up the following equations based on the given conditions:

1. The sum of the digits is 18:

2. The number obtained by interchanging the hundreds digit and the units digit ($100u + 10t + h$) is 594 less than the original number ($100h + 10t + u$).

This means the difference between the original number and the new number is 594:

Simplifying equation (ii):

Dividing both sides by 99:

3. The units digit is twice the hundreds digit:

Now we solve the system of equations:

$h + t + u = 18$ (i)

$h - u = 6$ (iii)

$u = 2h$ (iv)

Substitute the expression for $u$ from equation (iv) into equation (iii):

Substitute the value of $h$ into equation (iv) to find $u$:

Substitute the values of $h$ and $u$ into equation (i) to find $t$:

The values obtained for the digits are $h = -6$, $t = 36$, and $u = -12$.

By definition, the digits of a 3-digit number must be integers from 0 to 9, with the hundreds digit ($h$) being non-zero (i.e., $1 \le h \le 9$, $0 \le t \le 9$, $0 \le u \le 9$).

The calculated values $h = -6$, $t = 36$, and $u = -12$ do not satisfy these conditions as they are not valid digits.

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Conclusion:

The set of conditions provided in the problem leads to a system of equations whose solution for $h$, $t$, and $u$ are not valid digits. This indicates that there is an inconsistency in the problem statement, and therefore, no 3-digit number exists that simultaneously satisfies all the given conditions.

Let the three digits be A, B, and C. In a cryptarithmetic puzzle, each letter represents a unique digit from 0 to 9. Since A, B, and C are the first digits of 3-digit numbers, they cannot be zero.

The addition puzzle can be written as the sum of three numbers:

The first number is $100A + 10B + C$.

The second number is $100B + 10C + A$.

The third number is $100C + 10A + B$.

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We are given that the sum of these three numbers is 1665.

So, we have the equation:

Let's group the terms with A, B, and C:

Factor out 111 from the left side:

Now, divide both sides of equation (i) by 111 to find the sum of the digits A, B, and C:

Performing the division:

$\begin{array}{r} 15 \\ 111{\overline{\smash{\big)}\,1665}} \\ \underline{-~\phantom{(}111\phantom{-b)}} \\ 555 \\ \underline{-~\phantom{()}(555)} \\ 0\phantom{)} \end{array}$

So,

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Reasoning:

We need to find three distinct non-zero digits (A, B, and C) whose sum is 15.

Let's list possible combinations of three distinct non-zero digits that add up to 15:

• {1, 5, 9} ($1+5+9 = 15$)
• {1, 6, 8} ($1+6+8 = 15$)
• {2, 4, 9} ($2+4+9 = 15$)
• {2, 5, 8} ($2+5+8 = 15$)
• {2, 6, 7} ($2+6+7 = 15$)
• {3, 4, 8} ($3+4+8 = 15$)
• {3, 5, 7} ($3+5+7 = 15$)
• {4, 5, 6} ($4+5+6 = 15$)

Any set of three digits from this list can be assigned to A, B, and C in any order, and the sum $111(A+B+C)$ will always be 1665.

For example, let's take the set {4, 5, 6}. If we assign A=4, B=5, and C=6:

$\begin{array}{cccc} & 4 & 5 & 6 \\ + & 5 & 6 & 4 \\ + & 6 & 4 & 5 \\ \hline 1 & 6 & 6 & 5 \\ \hline \end{array}$

Checking the column sums:

Units column: $6 + 4 + 5 = 15$. Write down 5, carry over 1.

Tens column: $5 + 6 + 4 + 1 (\text{carry}) = 16$. Write down 6, carry over 1.

Hundreds column: $4 + 5 + 6 + 1 (\text{carry}) = 16$. Write down 6, carry over 1.

Thousands column: $1 (\text{carry}) = 1$.

The sum is indeed 1665.

This holds true for any permutation of the digits 4, 5, and 6 assigned to A, B, and C.

Similarly, any permutation of the digits from any of the other sets listed above would also satisfy the addition.

Since the puzzle structure is symmetric with respect to A, B, and C in the column sums, there isn't a unique assignment for A, B, and C individually based solely on the addition. However, a common convention for such puzzles is that *a* solution is expected.

Choosing the set {4, 5, 6}, one possible solution is to assign the values in increasing order.

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Solution:

The values for A, B, and C must be three distinct non-zero digits that sum to 15. One such set of digits is {4, 5, 6}.

One possible assignment for the values of A, B, and C is:

A = 4

B = 5

C = 6

(Any permutation of {4, 5, 6} is also a valid solution for the triplet (A, B, C)).