| Classwise Concept with Examples | ||||||
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| 6th | 7th | 8th | 9th | 10th | 11th | 12th |
| Content On This Page | ||
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| Numbers in General Form | Reversing the Digits of 2-Digit and 3-Digit Number | Letters for Digits |
| Number Puzzles and Games | Tests of Divisibility | |
Chapter 16 Playing with Numbers (Concepts)
Prepare to embark on a playful yet deeply insightful journey into the world of numbers with this chapter, aptly titled Playing with Numbers. Moving beyond routine calculations, we will explore the fascinating underlying structure, inherent patterns, and intriguing properties hidden within the number system itself. This exploration often feels more like solving puzzles than performing computations, aiming to develop your number sense, logical reasoning, and appreciation for the elegance of mathematical rules. We will learn to represent numbers in ways that reveal their secrets and discover shortcuts for determining divisibility without laborious calculations.
A cornerstone of this chapter is understanding the Generalized Form of numbers. While we typically write numbers using place value (e.g., 53 means 5 tens and 3 ones), representing them algebraically unlocks deeper understanding. A two-digit number, conventionally written as 'ab' (where 'a' is the tens digit and 'b' is the units digit, and $a \neq 0$), can be expressed in its generalized form as $10a + b$. Similarly, a three-digit number 'abc' (where $a \neq 0$) translates to $100a + 10b + c$. This algebraic representation is surprisingly powerful. It allows us to analyze and formally justify various number 'tricks' or puzzles. For instance, consider the puzzle of reversing the digits of a two-digit number and finding the difference: the difference between 'ab' and 'ba' is $(10a + b) - (10b + a) = 9a - 9b = 9(a-b)$. This immediately proves that the difference will always be divisible by $9$ (and also by $a-b$). Exploring such patterns using generalized forms is a key focus.
A major practical component involves mastering the Divisibility Rules. These rules are incredibly useful shortcuts that allow us to determine if a given number is exactly divisible by another specific number (typically small integers) without actually performing the long division. While some rules might be familiar (like those for 2, 5, and 10 based on the last digit), this chapter systematically introduces and often provides logical justifications (frequently using the generalized forms) for a broader set:
- Divisibility by 2: A number is divisible by 2 if its last digit is even (0, 2, 4, 6, or 8).
- Divisibility by 5: A number is divisible by 5 if its last digit is 0 or 5.
- Divisibility by 10: A number is divisible by 10 if its last digit is 0.
- Divisibility by 3: A number is divisible by 3 if the sum of its digits is divisible by 3. (e.g., for 123, $1+2+3=6$, which is divisible by 3, so 123 is divisible by 3).
- Divisibility by 9: Similar to 3, a number is divisible by 9 if the sum of its digits is divisible by 9. (e.g., for 729, $7+2+9=18$, which is divisible by 9, so 729 is divisible by 9).
- Divisibility by 4: A number is divisible by 4 if the number formed by its last two digits is divisible by 4. (e.g., 1736 is divisible by 4 because 36 is divisible by 4).
- Divisibility by 8: A number is divisible by 8 if the number formed by its last three digits is divisible by 8. (e.g., 51128 is divisible by 8 because 128 is divisible by 8).
- Divisibility by 6: A number is divisible by 6 if it is divisible by both 2 AND 3 (satisfying both rules simultaneously).
- Divisibility by 11: A number is divisible by 11 if the difference between the sum of the digits at odd places and the sum of the digits at even places (counting from the right) is either 0 or a multiple of 11. (e.g., for 13574, Odd places: $4+5+1=10$. Even places: $7+3=10$. Difference $10-10=0$. So 13574 is divisible by 11).
Applying these rules allows for rapid checks, especially with large numbers. Beyond rules, the chapter often incorporates engaging number games and logic puzzles like Cryptarithms. These are arithmetic problems where letters substitute for digits, and solving them requires a combination of logic, trial-and-error, and the application of number properties and divisibility rules learned. For example, solving SEND + MORE = MONEY requires finding the unique digit each letter represents. This type of problem-solving sharpens logical deduction and reinforces the chapter's core concepts in a fun, stimulating way, ultimately fostering a deeper number sense and appreciation for mathematical structure.
Numbers in General Form
In arithmetic, we are used to seeing numbers written using digits in their standard decimal form, like 25, 143, 5000, etc. The position of each digit in a number gives it a specific place value (units, tens, hundreds, thousands, and so on). We can express any number as the sum of the products of its digits and their corresponding place values. This way of writing numbers is called the general form or expanded form.
General Form of a 2-Digit Number
Let a 2-digit number be represented by 'ab'. It's important to understand that 'ab' here does not mean the product $a \times b$. Instead, it represents a number where 'a' is the digit in the tens place and 'b' is the digit in the units place.
The value of this number is determined by its place values:
Value of 'ab' $= (a \times 10) + (b \times 1)$
$= 10a + b$
... (i)
Important note: For a 2-digit number 'ab', the tens digit 'a' cannot be 0, otherwise it would become a 1-digit number. The units digit 'b' can be any digit from 0 to 9.
So, $a \in \{1, 2, ..., 9\}$ and $b \in \{0, 1, ..., 9\}$.
Example 1. Write the number 72 in its general form.
Answer:
The number is 72.
Here, the tens digit is $a = 7$ and the units digit is $b = 2$.
Using the general form $10a + b$:
General form of 72 $= 10 \times 7 + 2 = 70 + 2$.
Reversing the Digits of a 2-Digit Number
If the original number is 'ab' with value $10a + b$, then the number obtained by reversing its digits will be 'ba'.
In 'ba', the digit 'b' is now in the tens place and 'a' is in the units place.
Value of 'ba' $= (b \times 10) + (a \times 1)$
$= 10b + a$
... (ii)
General Form of a 3-Digit Number
Similarly, for a 3-digit number 'abc', 'a' is the hundreds digit, 'b' is the tens digit, and 'c' is the units digit.
The value of the number is given by:
Value of 'abc' $= (a \times 100) + (b \times 10) + (c \times 1)$
$= 100a + 10b + c$
... (iii)
Important note: For a 3-digit number 'abc', the hundreds digit 'a' cannot be 0. The digits 'b' and 'c' can be any digit from 0 to 9.
So, $a \in \{1, 2, ..., 9\}$ and $b, c \in \{0, 1, ..., 9\}$.
Example 2. Write the number 508 in its general form.
Answer:
The number is 508.
Here, the hundreds digit is $a=5$, the tens digit is $b=0$, and the units digit is $c=8$.
Using the general form $100a + 10b + c$:
General form of 508 $= 100 \times 5 + 10 \times 0 + 8 = 500 + 0 + 8$.
Reversing the Digits of a 3-Digit Number
If the original 3-digit number is 'abc' with a value of $100a + 10b + c$, the number obtained by reversing its digits will be 'cba'.
In the reversed number 'cba', the digit 'c' is in the hundreds place, 'b' remains in the tens place, and 'a' is now in the units place.
Value of 'cba' $= (c \times 100) + (b \times 10) + (a \times 1)$
$= 100c + 10b + a$
... (iv)
For 'cba' to remain a 3-digit number, the new hundreds digit 'c' cannot be 0.
Example 3. A number is given in its general form as $100 \times 9 + 10 \times 4 + 1 \times 3$. Find the number.
Answer:
The general form is $100 \times 9 + 10 \times 4 + 3$.
Comparing this with $100a + 10b + c$, we have:
Hundreds digit, $a = 9$.
Tens digit, $b = 4$.
Units digit, $c = 3$.
Therefore, the number is 943.
General Form for Numbers with More Digits
This pattern can be extended to numbers with any number of digits. For example, a 4-digit number 'abcd' has the value $1000a + 100b + 10c + d$, where 'a' cannot be zero.
Writing numbers in this general form is a powerful tool in algebra for exploring number properties and solving puzzles involving the digits of a number.
Reversing the Digits of 2-Digit and 3-Digit Numbers
In the previous section, we learned to represent numbers in a general form based on their place values. Using this general form, we can explore interesting patterns that arise when we manipulate the digits of numbers, such as reversing their order.
Reversing a 2-Digit Number
Let a 2-digit number be represented by its digits $a$ (tens digit) and $b$ (units digit). The number's value in general form is $10a + b$. For this to be a 2-digit number, $a$ must be a non-zero digit ($1 \le a \le 9$), and $b$ can be any digit ($0 \le b \le 9$).
Original number $= 10a + b$
When the digits are reversed, the new number has $b$ as the tens digit and $a$ as the units digit. Its value is $10b + a$.
Reversed number $= 10b + a$
Sum of a 2-Digit Number and its Reverse
Let's find the sum of the original number and its reverse:
Sum $= (10a + b) + (10b + a)$
Combining like terms (the 'a' terms and the 'b' terms):
$= (10a + a) + (b + 10b) = 11a + 11b$
Factoring out the common factor 11:
Sum $= 11(a+b)$
... (i)
Conclusion: The sum of any 2-digit number and the number formed by reversing its digits is always divisible by 11. The quotient is always the sum of the digits of the original number ($a+b$).
Difference of a 2-Digit Number and its Reverse
Now, let's find the difference between the original number and its reverse. Let's assume the original number is greater than the reversed number (i.e., $a > b$).
Difference $= (10a + b) - (10b + a)$
Removing the parentheses and combining like terms:
$= 10a + b - 10b - a = (10a - a) + (b - 10b)$
$= 9a - 9b$
Factoring out the common factor 9:
Difference $= 9(a-b)$
... (ii)
Conclusion: The difference between any 2-digit number and the number formed by reversing its digits is always divisible by 9. The quotient is always the difference between the digits of the original number ($a-b$).
Example 1. Consider the number 62. Find the sum and difference with its reversed number and verify the properties.
Answer:
Original number = 62. Here, the tens digit is $a=6$ and the units digit is $b=2$.
Reversed number = 26.
Verification of the Sum Property:
Sum $= 62 + 26 = 88$
To check if it's divisible by 11, we divide: $88 \div 11 = 8$.
The sum of the digits of the original number is $a+b = 6+2 = 8$.
The quotient (8) matches the sum of the digits (8), as expected from the formula $11(a+b)$.
Verification of the Difference Property:
Difference $= 62 - 26 = 36$
To check if it's divisible by 9, we divide: $36 \div 9 = 4$.
The difference between the digits of the original number is $a-b = 6-2 = 4$.
The quotient (4) matches the difference of the digits (4), as expected from the formula $9(a-b)$.
Reversing a 3-Digit Number
Let the original 3-digit number be 'abc'. In general form, its value is $100a + 10b + c$. For this to be a 3-digit number, $a \neq 0$.
Original number $= 100a + 10b + c$
When the digits are reversed, the new number is 'cba'. Its value is $100c + 10b + a$. For the reversed number to be a 3-digit number, $c \neq 0$.
Reversed number $= 100c + 10b + a$
Difference of a 3-Digit Number and its Reverse
Let's find the difference between the original number and its reverse. Assume $a > c$.
Difference $= (100a + 10b + c) - (100c + 10b + a)$
Removing the parentheses and combining like terms:
$= 100a + 10b + c - 100c - 10b - a$
$= (100a - a) + (10b - 10b) + (c - 100c)$
$= 99a + 0 - 99c = 99a - 99c$
Factoring out the common factor 99:
Difference $= 99(a-c)$
... (iii)
Conclusion: The difference between any 3-digit number and the number formed by reversing its digits is always divisible by 99. The quotient is always the difference between the hundreds and the units digits ($a-c$).
Example 2. Take the number 734. Find the difference with its reversed number and verify the property.
Answer:
Original number = 734. Here, the digits are $a=7$ (hundreds), $b=3$ (tens), and $c=4$ (units).
Reversed number = 437.
Verification of the Difference Property:
Difference $= 734 - 437$
Let's perform the subtraction:
$\begin{array}{cc} & \phantom{}^{6}\cancel{7} & \phantom{}^{12}\cancel{3} & \phantom{}^{14}\cancel{4} \\ - & 4 & 3 & 7 \\ \hline & 2 & 9 & 7 \\ \hline \end{array}$The difference is 297.
To check if it's divisible by 99, we divide: $297 \div 99 = 3$.
The difference between the hundreds and units digits of the original number is $a-c = 7-4 = 3$.
The quotient (3) matches the difference of the digits (3), as expected from the formula $99(a-c)$.
Letters for Digits
In puzzles and recreational mathematics, you sometimes see arithmetic problems where the digits are replaced by letters. Each letter represents a unique digit from 0 to 9. The goal is to decode the puzzle by finding which digit each letter stands for. These puzzles are also known as cryptarithmetic problems or cryptarithms.
Solving these puzzles requires a combination of basic arithmetic rules and logical deduction. By carefully examining the structure of the problem, we can uncover the hidden values.
Rules for Solving Letter Puzzles
When solving puzzles where letters represent digits, you must follow these fundamental rules:
- One Letter, One Digit: Each letter in the puzzle stands for exactly one digit from 0 to 9. If a letter appears multiple times, it must represent the same digit every time.
- One Digit, One Letter: Each digit from 0 to 9 can be represented by at most one letter. If two different letters appear (e.g., A and B), they must represent different digits.
- No Leading Zeros: The first digit of any number in the puzzle cannot be zero. For example, in the number 'AB', A cannot be 0.
The strategy is to start with the column that gives the most information (often the units column) and use logical steps, considering carry-overs or borrowing, to find the value of each letter.
Examples of Letter Puzzles
Example 1. Find the values of the letters A and B in the following addition:
$\begin{array}{@{}c@{}c@{}c} & 3 & A \\ {} + {} & 2 & 5 \\ \hline & B & 2 \end{array}$Answer:
We need to solve the addition problem:
$\begin{array}{@{}c@{}c@{}c} & 3 & A \\ {} + {} & 2 & 5 \\ \hline & B & 2 \end{array}$Step 1: Analyse the Units Column.
In the rightmost column, we have $A + 5$. The result ends in the digit 2. This means $A+5$ must be a number like 2, 12, 22, etc.
Since A is a single digit (from 0 to 9), the maximum value of $A+5$ is $9+5=14$. Therefore, the only possibility is that $A+5$ equals 12.
$A + 5 = 12$
Solving for A, we get $A = 12 - 5 = 7$.
So, A = 7. Because $7+5=12$, we write down 2 in the units place of the sum and carry over 1 to the tens column.
Step 2: Analyse the Tens Column.
In the tens column, we add the digits 3, 2, and the carry-over from the units column, which is 1. The result of this sum is the digit B.
$B = 3 + 2 + 1_{\text{(carry)}}$
$B = 6$
So, B = 6. Since B is the first digit of the result 'B2', it cannot be 0. Our value B=6 satisfies this condition.
Step 3: Verify the Solution.
Let's substitute A=7 and B=6 back into the puzzle:
$\begin{array}{@{}c@{\,}c@{}c} & 3 & 7 \\ {} + {} & 2 & 5 \\ \hline & 6 & 2 \end{array}$The addition $37 + 25 = 62$ is correct. Therefore, the solution is A = 7 and B = 6.
Example 2. Find the digit represented by A in the following multiplication:
$\begin{array}{@{}c@{}c@{}c@{}c} & & 1 & A \\ {} \times {} & & & A \\ \hline & & 9 & A \end{array}$Answer:
The puzzle is the multiplication of a 2-digit number '1A' by the single digit 'A', resulting in the 2-digit number '9A'.
Step 1: Analyse the Units Column.
The units digit of the product is determined by the units digit of $A \times A$. In this puzzle, the units digit of the product is also A. So, we are looking for a digit A such that the number $A \times A$ (or $A^2$) ends with the digit A.
Let's test the possible digits for A:
- $0 \times 0 = 0$. (Possible, if A=0). But if A=0, the problem is $10 \times 0 = 0$. The result should be '9A' or 90. $0 \neq 90$. So, A cannot be 0.
- $1 \times 1 = 1$. (Possible, if A=1). Let's check: $11 \times 1 = 11$. The result should be '9A' or 91. $11 \neq 91$. So, A cannot be 1.
- $5 \times 5 = 25$. (Possible, if A=5). Let's check: $15 \times 5 = 75$. The result should be '9A' or 95. $75 \neq 95$. So, A cannot be 5.
- $6 \times 6 = 36$. (Possible, if A=6). Let's check: $16 \times 6 = 96$. The result should be '9A' or 96. This matches perfectly!
The only remaining possibility from the units digit analysis is that A=6.
Step 2: Verify the Full Multiplication.
Let's perform the multiplication with A=6:
$\begin{array}{@{}c@{}c@{}c@{}c} & & 1 & 6 \\ {} \times {} & & & 6 \\ \hline & & 9 & 6 \end{array}$- Units place: $6 \times 6 = 36$. Write down 6, carry over 3.
- Tens place: $6 \times 1 = 6$. Add the carry-over: $6 + 3 = 9$. Write down 9.
The result is 96, which matches the format '9A' with A=6. The solution is correct.
The digit represented by A is 6.
Example 3. Find the digits A and B in the following addition:
$\begin{array}{@{}c@{\,}c} & A & B \\ {} + {} & 3 & 7 \\ \hline & 6 & A \end{array}$Answer:
The puzzle is the addition of 'AB' and 37 to get '6A'. A cannot be 0 since it is the first digit of 'AB'.
Step 1: Analyse the Tens Column.
In the tens column, we have $A + 3 + (\text{carry from units}) = 6$.
The carry from the units column can only be 0 or 1. Let's test these two cases.
- Case 1: The carry is 0. If there is no carry, then $A + 3 + 0 = 6$, which means $A=3$.
- Case 2: The carry is 1. If there is a carry of 1, then $A + 3 + 1 = 6$, which means $A=2$.
So, A can only be 2 or 3. We can now use the units column to find out which case is correct.
Step 2: Analyse the Units Column using the possibilities for A.
The units column equation is $B + 7 =$ a number ending in A. This is where the carry comes from.
- Test Case 1 (A=3): If A=3, then $B+7$ must end in 3. This means $B+7 = 3$ (impossible as B is not negative) or $B+7 = 13$. If $B+7=13$, then $B = 13-7=6$. This would generate a carry-over of 1 to the tens column. But our assumption for Case 1 was that the carry is 0. This is a contradiction. So, A cannot be 3.
- Test Case 2 (A=2): If A=2, then $B+7$ must end in 2. This means $B+7=12$. If $B+7=12$, then $B=12-7=5$. This generates a carry-over of 1 to the tens column. Our assumption for Case 2 was that the carry is 1. This is consistent.
Step 3: State and Verify the Solution.
The only consistent solution is A=2 and B=5.
Let's check the original addition:
$\begin{array}{@{}c@{\,}c} & 2 & 5 \\ {} + {} & 3 & 7 \\ \hline & 6 & 2 \end{array}$The addition $25 + 37 = 62$ is correct. The result '6A' matches '62'.
The digits are A=2 and B=5.
Number Puzzles and Games
The properties of numbers written in general form and the relationships we discovered when reversing digits can be used to create interesting number puzzles and games. These puzzles are not only fun but also help reinforce the understanding of place value and algebraic representation of numbers.
Using Properties to Create Puzzles
Consider the patterns we observed:
- The sum of a 2-digit number ($10a+b$) and its reverse ($10b+a$) is $11(a+b)$. This sum is always a multiple of 11, and the quotient is the sum of the digits.
- The difference between a 2-digit number ($10a+b$) and its reverse ($10b+a$) is $9(a-b)$ or $9(b-a)$. The absolute difference is $9|a-b|$. This difference is always a multiple of 9, and the quotient is the absolute difference of the digits.
- The difference between a 3-digit number ($100a+10b+c$) and its reverse ($100c+10b+a$) is $99(a-c)$ or $99(c-a)$. The absolute difference is $99|a-c|$. This difference is always a multiple of 99, and the quotient is the absolute difference of the hundreds and units digits.
We can use these predictable outcomes to create mind-reading tricks or puzzles.
Example 1. A person is asked to perform the following steps:
- Think of any 2-digit number.
- Reverse the digits of the number.
- Add the original number and the reversed number.
If the person tells you the final sum is 110, how can you determine the sum of the digits of the number they originally thought of?
Answer:
This puzzle works because the sum of a 2-digit number and its reverse always follows a specific mathematical pattern.
Let the original 2-digit number be 'ab'. In general form, its value is $10a + b$.
The number obtained by reversing the digits is 'ba'. In general form, its value is $10b + a$.
Now, let's find the sum of these two numbers:
Sum $= (10a + b) + (10b + a)$
By combining like terms:
Sum $= 11a + 11b = 11(a+b)$
This formula shows that the final sum is always 11 times the sum of the digits ($a+b$). Therefore, to find the sum of the digits, we just need to divide the final sum by 11.
The Solution:
The person's final sum is 110. To find the sum of the digits of their original number, we perform the calculation:
Sum of digits $= \frac{\text{Final Sum}}{11} = \frac{110}{11} = 10$
You can instantly tell them that the sum of the digits of the number they thought of was 10. (For example, if they chose 73, the reverse is 37. Sum = $73+37=110$. Sum of digits = $7+3=10$.)
Example 2. A person is asked to perform the following steps:
- Think of any 2-digit number where the two digits are different.
- Reverse the digits of the number.
- Subtract the smaller number from the larger number.
If the person tells you the final difference is 27, how can you determine the difference between the digits of their original number?
Answer:
This puzzle uses the property of the difference between a 2-digit number and its reverse.
Let the original number be 'ab', with value $10a+b$. The reversed number is 'ba', with value $10b+a$. Let's assume $a > b$, so the original number is larger.
The difference is:
Difference $= (10a + b) - (10b + a) = 9a - 9b = 9(a-b)$
This formula shows that the difference is always 9 times the difference of the digits ($a-b$). So, to find the difference between the digits, we can divide the final result by 9.
The Solution:
The person's final difference is 27. To find the difference between the digits of their original number, we calculate:
Difference of digits $= \frac{\text{Final Difference}}{9} = \frac{27}{9} = 3$
You can state that the difference between the digits of the number they thought of was 3. (For example, if they chose 52, the reverse is 25. Difference = $52-25=27$. Difference of digits = $5-2=3$.)
Example 3. A person follows these steps:
- Think of any 3-digit number where the hundreds digit and units digit are different.
- Reverse the digits of the number.
- Subtract the smaller number from the larger number.
If the final difference is 396, what can you determine about the digits of their original number?
Answer:
This puzzle relies on the property of the difference between a 3-digit number and its reverse.
Let the original number be 'abc', with a value of $100a+10b+c$. The reversed number is 'cba', with a value of $100c+10b+a$. Assume $a > c$.
The difference is:
Difference $= (100a + 10b + c) - (100c + 10b + a)$
$ = 99a - 99c = 99(a-c)$
The formula shows that the difference is always 99 times the difference between the hundreds and units digits ($a-c$).
The Solution:
The person's final difference is 396. To find the difference between the first and last digits, we divide this result by 99.
Difference of hundreds and units digits $= \frac{\text{Final Difference}}{99} = \frac{396}{99} = 4$
You can tell them that the difference between the first and last digits of the number they thought of was 4. (For example, if they chose 854, the reverse is 458. Difference = $854-458=396$. Difference of first and last digits = $8-4=4$.)
Tests of Divisibility
Divisibility tests are shortcuts that allow us to determine if a number can be divided evenly by another number without performing the actual division. In this section, we will review the common divisibility tests and use the general form of numbers to understand why they work.
Divisibility by 2, 5, and 10
These tests are the simplest because they only depend on the last digit (the units digit) of the number.
Consider any number, for example, a 3-digit number 'abc', whose value is $100a + 10b + c$. We can rewrite this as $10(10a+b) + c$. Since the first part, $10(10a+b)$, is always a multiple of 10, it is also a multiple of 2 and 5. Therefore, the divisibility of the entire number by 2, 5, or 10 depends solely on the units digit, 'c'.
1. Divisibility by 2
Rule: A number is divisible by 2 if its units digit is an even number (0, 2, 4, 6, or 8).
Example: 548 is divisible by 2 because its units digit is 8, which is even.
Example: 1,293 is not divisible by 2 because its units digit is 3, which is odd.
2. Divisibility by 5
Rule: A number is divisible by 5 if its units digit is either 0 or 5.
Example: 3,475 is divisible by 5 because its units digit is 5.
Example: 860 is divisible by 5 because its units digit is 0.
3. Divisibility by 10
Rule: A number is divisible by 10 if its units digit is 0.
Example: 7,290 is divisible by 10 because it ends in 0.
Example: 451 is not divisible by 10.
Divisibility by 3 and 9
These tests depend on the sum of the digits of the number.
Let's see why this works for a 3-digit number 'abc' = $100a + 10b + c$. We can rewrite it as:
$100a + 10b + c = (99a + a) + (9b + b) + c$
$ = (99a + 9b) + (a+b+c)$
The part $(99a + 9b)$ is always divisible by both 3 and 9. Therefore, the divisibility of the original number by 3 or 9 depends entirely on the divisibility of the sum of its digits, $(a+b+c)$.
4. Divisibility by 3
Rule: A number is divisible by 3 if the sum of its digits is divisible by 3.
Example: Is 528 divisible by 3?
Sum of digits = $5+2+8 = 15$. Since 15 is divisible by 3 ($15 \div 3 = 5$), the number 528 is also divisible by 3.
5. Divisibility by 9
Rule: A number is divisible by 9 if the sum of its digits is divisible by 9.
Example: Is 459 divisible by 9?
Sum of digits = $4+5+9 = 18$. Since 18 is divisible by 9 ($18 \div 9 = 2$), the number 459 is also divisible by 9.
Note: Any number divisible by 9 is also divisible by 3, but the reverse is not always true (e.g., 12 is divisible by 3 but not by 9).
6. Divisibility by 6
Rule: A number is divisible by 6 if it is divisible by both 2 and 3.
Example: Is 216 divisible by 6?
- Check for divisibility by 2: The units digit is 6 (even), so it is divisible by 2.
- Check for divisibility by 3: The sum of digits is $2+1+6 = 9$. Since 9 is divisible by 3, the number is divisible by 3.
Since 216 is divisible by both 2 and 3, it is divisible by 6.
Divisibility by 11
This test involves finding the difference between the sum of digits at odd places and the sum of digits at even places, starting from the right (units place).
Rule: A number is divisible by 11 if the difference between the sum of its digits in odd places (from the right) and the sum of its digits in even places (from the right) is either 0 or a multiple of 11.
Let's see why this works for a 4-digit number 'abcd' = $1000a + 100b + 10c + d$. We can write it as:
$(1001a - a) + (99b + b) + (11c - c) + d$
$= (1001a + 99b + 11c) + (-a + b - c + d)$
$= (\text{A multiple of 11}) + (b+d) - (a+c)$
The divisibility depends on whether $(b+d) - (a+c)$ is divisible by 11. Here, 'd' and 'b' are in odd places (1st and 3rd from right), and 'c' and 'a' are in even places (2nd and 4th from right). Ah wait, I messed up the order. Let's fix that.
Places from right: d(1st), c(2nd), b(3rd), a(4th).
Sum of odd-placed digits = $d+b$. Sum of even-placed digits = $c+a$.
The difference is $(d+b) - (c+a)$. The logic holds.
Example: Is 61809 divisible by 11?
The number is 6 1 8 0 9.
Sum of digits at odd places (from right: 9, 8, 6) = $9 + 8 + 6 = 23$.
Sum of digits at even places (from right: 0, 1) = $0 + 1 = 1$.
Difference = $23 - 1 = 22$.
Since 22 is a multiple of 11, the number 61809 is divisible by 11.
Example: Is 1331 divisible by 11?
The number is 1 3 3 1.
Sum of digits at odd places (1, 3) = $1 + 3 = 4$.
Sum of digits at even places (3, 1) = $3 + 1 = 4$.
Difference = $4 - 4 = 0$.
Since the difference is 0, the number 1331 is divisible by 11.