Chapter 6 Squares and Square Roots

The unit digit of the square of a number depends only on the unit digit of the original number. To find the unit digit of the square of each number, we need to square the unit digit of the given number and look at the unit digit of the result.

Let's find the unit digit of the square for each number:

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(i) 81

The unit digit of 81 is 1.

The square of the unit digit is $1^2 = 1$.

So, the unit digit of the square of 81 is 1.

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(ii) 272

The unit digit of 272 is 2.

The square of the unit digit is $2^2 = 4$.

So, the unit digit of the square of 272 is 4.

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(iii) 799

The unit digit of 799 is 9.

The square of the unit digit is $9^2 = 81$.

The unit digit of 81 is 1.

So, the unit digit of the square of 799 is 1.

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(iv) 3853

The unit digit of 3853 is 3.

The square of the unit digit is $3^2 = 9$.

So, the unit digit of the square of 3853 is 9.

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(v) 1234

The unit digit of 1234 is 4.

The square of the unit digit is $4^2 = 16$.

The unit digit of 16 is 6.

So, the unit digit of the square of 1234 is 6.

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(vi) 26387

The unit digit of 26387 is 7.

The square of the unit digit is $7^2 = 49$.

The unit digit of 49 is 9.

So, the unit digit of the square of 26387 is 9.

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(vii) 52698

The unit digit of 52698 is 8.

The square of the unit digit is $8^2 = 64$.

The unit digit of 64 is 4.

So, the unit digit of the square of 52698 is 4.

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(viii) 99880

The unit digit of 99880 is 0.

The square of the unit digit is $0^2 = 0$.

So, the unit digit of the square of 99880 is 0.

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(ix) 12796

The unit digit of 12796 is 6.

The square of the unit digit is $6^2 = 36$.

The unit digit of 36 is 6.

So, the unit digit of the square of 12796 is 6.

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(x) 55555

The unit digit of 55555 is 5.

The square of the unit digit is $5^2 = 25$.

The unit digit of 25 is 5.

So, the unit digit of the square of 55555 is 5.

Reason: We know that perfect squares can only end with the digits 0, 1, 4, 5, 6, or 9. Also, a number ending in zeros is a perfect square only if it ends with an even number of zeros.

Numbers ending in the digits 2, 3, 7, or 8 are never perfect squares.

Numbers ending in an odd number of zeros are never perfect squares.

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(i) 1057

The unit digit of 1057 is 7.

Since perfect squares cannot end in 7, 1057 is not a perfect square.

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(ii) 23453

The unit digit of 23453 is 3.

Since perfect squares cannot end in 3, 23453 is not a perfect square.

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(iii) 7928

The unit digit of 7928 is 8.

Since perfect squares cannot end in 8, 7928 is not a perfect square.

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(iv) 222222

The unit digit of 222222 is 2.

Since perfect squares cannot end in 2, 222222 is not a perfect square.

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(v) 64000

This number ends in three zeros.

Since the number of trailing zeros is 3 (which is odd), 64000 is not a perfect square.

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(vi) 89722

The unit digit of 89722 is 2.

Since perfect squares cannot end in 2, 89722 is not a perfect square.

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(vii) 222000

This number ends in three zeros.

Since the number of trailing zeros is 3 (which is odd), 222000 is not a perfect square.

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(viii) 505050

This number ends in one zero.

Since the number of trailing zeros is 1 (which is odd), 505050 is not a perfect square.

We need to determine which of the given numbers, when squared, will result in an odd number.

Rule: The square of an odd number is always an odd number, and the square of an even number is always an even number.

A number is odd if its unit digit is 1, 3, 5, 7, or 9.

A number is even if its unit digit is 0, 2, 4, 6, or 8.

We will check the unit digit of each given number to determine if the number is odd or even.

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(i) 431

The unit digit of 431 is 1.

Since the unit digit is 1, the number 431 is an odd number.

The square of an odd number is odd.

Therefore, the square of 431 will be an odd number.

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(ii) 2826

The unit digit of 2826 is 6.

Since the unit digit is 6, the number 2826 is an even number.

The square of an even number is even.

Therefore, the square of 2826 will be an even number.

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(iii) 7779

The unit digit of 7779 is 9.

Since the unit digit is 9, the number 7779 is an odd number.

The square of an odd number is odd.

Therefore, the square of 7779 will be an odd number.

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(iv) 82004

The unit digit of 82004 is 4.

Since the unit digit is 4, the number 82004 is an even number.

The square of an even number is even.

Therefore, the square of 82004 will be an even number.

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Conclusion:

The squares of the following numbers would be odd numbers:

(i) 431

(iii) 7779

Let's observe the given pattern:

$11^2 = 121$

$101^2 = 10201$

$1001^2 = 1002001$

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We can see a relationship between the number being squared and the result:

• The number being squared is of the form $1$ followed by $n$ zeros, followed by $1$.
• The result is $1$ followed by $n$ zeros, followed by $2$, followed by $n$ zeros, followed by $1$.

Let's check this pattern:

• For $11^2$: $n=0$ zeros. Result is $1 \underbrace{}_{0 \text{ zeros}} 2 \underbrace{}_{0 \text{ zeros}} 1 = 121$. This matches.
• For $101^2$: $n=1$ zero. Result is $1 \underbrace{0}_{1 \text{ zero}} 2 \underbrace{0}_{1 \text{ zero}} 1 = 10201$. This matches.
• For $1001^2$: $n=2$ zeros. Result is $1 \underbrace{00}_{2 \text{ zeros}} 2 \underbrace{00}_{2 \text{ zeros}} 1 = 1002001$. This matches.

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Now we apply this pattern to find the missing digits:

For $100001^2$:

The number $100001$ has $n=4$ zeros between the ones.

According to the pattern, the square should be $1$ followed by 4 zeros, then $2$, then 4 zeros, then $1$.

$100001^2 = 1 \underbrace{0000}_{4 \text{ zeros}} 2 \underbrace{0000}_{4 \text{ zeros}} 1 = 10000200001$.

Comparing this with $1 ......... 2 ......... 1$, the missing digits are 0000 in the first blank and 0000 in the second blank.

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For $10000001^2$:

The number $10000001$ has $n=6$ zeros between the ones.

According to the pattern, the square should be $1$ followed by 6 zeros, then $2$, then 6 zeros, then $1$.

$10000001^2 = 1 \underbrace{000000}_{6 \text{ zeros}} 2 \underbrace{000000}_{6 \text{ zeros}} 1 = 100000020000001$.

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The completed pattern is:

Let's analyze the given pattern:

$11^2 = 121$

$101^2 = 10201$

$10101^2 = 102030201$

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Observation:

• The numbers being squared consist of alternating 1s and 0s, starting and ending with 1.
• The number of 1s in the base number corresponds to the peak digit in the squared result.
• The squared result consists of digits increasing from 1 up to the number of 1s in the base, and then decreasing back to 1, with a '0' inserted between each consecutive digit.

Let $k$ be the number of 1s in the base number.

• For $11^2$: $k=2$. Result is $121$. (Digits go up to 2 and down)
• For $101^2$: $k=2$. Result is $10201$. (Digits go up to 2 and down, separated by 0)
• For $10101^2$: $k=3$. Result is $102030201$. (Digits go up to 3 and down, separated by 0)

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Applying the pattern to find the missing numbers:

1. $1010101^2 = ?$

The base number $1010101$ has four 1s ($k=4$).

Following the pattern, the result should have digits going up to 4 and back down to 1, separated by zeros: $1020304030201$.

So, $1010101^2 = \mathbf{1020304030201}$.

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2. $?^2 = 10203040504030201$

The result $10203040504030201$ has digits going up to 5 and back down to 1, separated by zeros.

This means the base number must have five 1s ($k=5$).

The base number should consist of five 1s alternating with zeros, starting and ending with 1.

So, the missing base number is $\mathbf{101010101}$.

$(\mathbf{101010101})^2 = 10203040504030201$.

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The completed pattern is:

Let's observe the pattern in the given equations:

$1^2 + 2^2 + 2^2 = 3^2$

$2^2 + 3^2 + 6^2 = 7^2$

$3^2 + 4^2 + 12^2 = 13^2$

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We can identify a relationship between the numbers in each row:

• The second number is one more than the first number. (e.g., $2 = 1+1$, $3 = 2+1$, $4 = 3+1$)
• The third number is the product of the first two numbers. (e.g., $2 = 1 \times 2$, $6 = 2 \times 3$, $12 = 3 \times 4$)
• The number on the right side of the equation is one more than the third number. (e.g., $3 = 2+1$, $7 = 6+1$, $13 = 12+1$)

In general, if the first number is $n$, the pattern follows the equation:

$n^2 + (n+1)^2 + [n(n+1)]^2 = [n(n+1) + 1]^2$

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Now we use this pattern to find the missing numbers:

Row 4: $4^2 + 5^2 + \underline{\hspace{0.5cm}}^2 = 21^2$

Here, the first number is $n=4$.

The second number is $n+1 = 4+1 = 5$. (Matches)

The third number should be $n(n+1) = 4 \times (4+1) = 4 \times 5 = 20$.

The number on the right side should be $n(n+1) + 1 = 20 + 1 = 21$. (Matches)

So, the missing number is 20.

$4^2 + 5^2 + \mathbf{20}^2 = 21^2$

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Row 5: $5^2 + \underline{\hspace{0.5cm}}^2 + 30^2 = 31^2$

Here, the first number is $n=5$.

The second number should be $n+1 = 5+1 = 6$.

The third number is $n(n+1) = 5 \times (5+1) = 5 \times 6 = 30$. (Matches)

The number on the right side is $n(n+1) + 1 = 30 + 1 = 31$. (Matches)

So, the missing number is 6.

$5^2 + \mathbf{6}^2 + 30^2 = 31^2$

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Row 6: $6^2 + 7^2 + \underline{\hspace{0.5cm}}^2 = \underline{\hspace{0.5cm}}^2$

Here, the first number is $n=6$.

The second number is $n+1 = 6+1 = 7$. (Matches)

The third number should be $n(n+1) = 6 \times (6+1) = 6 \times 7 = 42$.

The number on the right side should be $n(n+1) + 1 = 42 + 1 = 43$.

So, the missing numbers are 42 and 43.

$6^2 + 7^2 + \mathbf{42}^2 = \mathbf{43}^2$

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The completed pattern with missing numbers filled in:

$1^2 + 2^2 + 2^2 = 3^2$

$2^2 + 3^2 + 6^2 = 7^2$

$3^2 + 4^2 + 12^2 = 13^2$

$4^2 + 5^2 + \mathbf{20}^2 = 21^2$

$5^2 + \mathbf{6}^2 + 30^2 = 31^2$

$6^2 + 7^2 + \mathbf{42}^2 = \mathbf{43}^2$

We know that the sum of the first $n$ odd natural numbers is given by the formula $n^2$. We will use this property to find the sums without performing the addition.

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(i) 1 + 3 + 5 + 7 + 9

These are the first 5 odd natural numbers.

Here, $n = 5$.

Therefore, the sum is $n^2 = 5^2$.

Sum = $5 \times 5 = \mathbf{25}$.

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(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19

These are the first 10 odd natural numbers.

Here, $n = 10$.

Therefore, the sum is $n^2 = 10^2$.

Sum = $10 \times 10 = \mathbf{100}$.

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(iii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23

These are the first 12 odd natural numbers.

Here, $n = 12$.

Therefore, the sum is $n^2 = 12^2$.

Sum = $12 \times 12 = \mathbf{144}$.

We know that any perfect square $n^2$ can be expressed as the sum of the first $n$ consecutive odd natural numbers.

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(i) Express 49 as the sum of 7 odd numbers.

The given number is 49.

We know that $49 = 7^2$.

Therefore, 49 is the sum of the first 7 odd natural numbers.

The first 7 odd natural numbers are 1, 3, 5, 7, 9, 11, 13.

So, we can express 49 as:

$\mathbf{49 = 1 + 3 + 5 + 7 + 9 + 11 + 13}$

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(ii) Express 121 as the sum of 11 odd numbers.

The given number is 121.

We know that $121 = 11^2$.

Therefore, 121 is the sum of the first 11 odd natural numbers.

The first 11 odd natural numbers are 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21.

So, we can express 121 as:

$\mathbf{121 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21}$

We need to find the number of non-square integers that lie between the squares of two consecutive integers, $n$ and $(n+1)$.

The squares are $n^2$ and $(n+1)^2$.

The number of integers strictly between $n^2$ and $(n+1)^2$ is given by the difference between the squares minus 1:

Number of integers = $(n+1)^2 - n^2 - 1$

Expanding $(n+1)^2$: $(n+1)^2 = n^2 + 2n + 1$.

Substituting this back:

Number of integers = $(n^2 + 2n + 1) - n^2 - 1$

Number of integers = $n^2 + 2n + 1 - n^2 - 1 = 2n$.

So, there are $2n$ non-square integers between the squares of the consecutive integers $n$ and $(n+1)$.

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(i) 12 and 13

Here, $n=12$ and $n+1=13$.

The number of integers lying between $12^2$ and $13^2$ is $2n$.

Number of integers = $2 \times 12 = \mathbf{24}$.

(Check: $12^2 = 144$, $13^2 = 169$. The integers are 145, 146, ..., 168. Count = $168 - 144 = 24$.)

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(ii) 25 and 26

Here, $n=25$ and $n+1=26$.

The number of integers lying between $25^2$ and $26^2$ is $2n$.

Number of integers = $2 \times 25 = \mathbf{50}$.

(Check: $25^2 = 625$, $26^2 = 676$. The integers are 626, 627, ..., 675. Count = $675 - 625 = 50$.)

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(iii) 99 and 100

Here, $n=99$ and $n+1=100$.

The number of integers lying between $99^2$ and $100^2$ is $2n$.

Number of integers = $2 \times 99 = \mathbf{198}$.

(Check: $99^2 = 9801$, $100^2 = 10000$. The integers are 9802, 9803, ..., 9999. Count = $9999 - 9801 = 198$.)

We can find the square of the numbers by expressing them as a sum or difference of two numbers and using the algebraic identities:

$(a+b)^2 = a^2 + 2ab + b^2$

$(a-b)^2 = a^2 - 2ab + b^2$

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(i) 39

We can write 39 as $(40 - 1)$.

So, $39^2 = (40 - 1)^2$.

Using the identity $(a-b)^2 = a^2 - 2ab + b^2$, with $a=40$ and $b=1$:

$39^2 = 40^2 - 2(40)(1) + 1^2$

$39^2 = 1600 - 80 + 1$

$39^2 = 1520 + 1$

$39^2 = \mathbf{1521}$

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(ii) 42

We can write 42 as $(40 + 2)$.

So, $42^2 = (40 + 2)^2$.

Using the identity $(a+b)^2 = a^2 + 2ab + b^2$, with $a=40$ and $b=2$:

$42^2 = 40^2 + 2(40)(2) + 2^2$

$42^2 = 1600 + 160 + 4$

$42^2 = 1760 + 4$

$42^2 = \mathbf{1764}$

Solution:

A Pythagorean triplet consists of three positive integers a, b, and c, such that $a^2 + b^2 = c^2$.

For any natural number $m > 1$, the triplet $(2m, m^2 - 1, m^2 + 1)$ forms a Pythagorean triplet.

We are given that the smallest member of the triplet is 8. We will check which member of the general form can be equal to 8.

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Case 1: Let $m^2 - 1 = 8$

If $m^2 - 1 = 8$, then $m^2 = 8 + 1 = 9$.

This gives $m = \sqrt{9} = 3$. Since m is a natural number greater than 1, this is a valid value.

Let's find the other two members of the triplet:

$2m = 2 \times 3 = 6$

$m^2 + 1 = 3^2 + 1 = 9 + 1 = 10$

The triplet is (6, 8, 10). In this triplet, the smallest member is 6, not 8. Therefore, this case does not satisfy the condition of the question.

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Case 2: Let $2m = 8$

If $2m = 8$, then $m = \frac{8}{2} = 4$. Since m is a natural number greater than 1, this is a valid value.

Let's find the other two members of the triplet:

$m^2 - 1 = 4^2 - 1 = 16 - 1 = 15$

$m^2 + 1 = 4^2 + 1 = 16 + 1 = 17$

The triplet is (8, 15, 17). Here, the smallest member is 8. This case satisfies the condition of the question.

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Verification:

To verify if (8, 15, 17) is a Pythagorean triplet, we check if the sum of the squares of the two smaller numbers equals the square of the largest number.

$8^2 + 15^2 = 64 + 225 = 289$

$17^2 = 289$

Since $8^2 + 15^2 = 17^2$, the triplet is a valid Pythagorean triplet.

Therefore, the Pythagorean triplet whose smallest member is 8 is (8, 15, 17).

Solution:

We know that for any natural number $m > 1$, the set of numbers $(2m, m^2 - 1, m^2 + 1)$ forms a Pythagorean triplet.

We are given that one member of the triplet is 12. We will try to find a suitable value of 'm' by equating each member of the general form to 12.

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Case 1: Let $m^2 - 1 = 12$

If we set $m^2 - 1 = 12$, then we get:

$m^2 = 12 + 1$

$m^2 = 13$

$m = \sqrt{13}$

Since the value of $m$ is not an integer, this case does not give us a Pythagorean triplet with integer members.

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Case 2: Let $2m = 12$

If we set $2m = 12$, then we get:

$m = \frac{12}{2} = 6$

Since $m = 6$ is an integer greater than 1, we can use this value to find the other two members of the triplet.

The first member is $2m = 12$.

The second member is $m^2 - 1$:

$m^2 - 1 = (6)^2 - 1 = 36 - 1 = 35$

The third member is $m^2 + 1$:

$m^2 + 1 = (6)^2 + 1 = 36 + 1 = 37$

Thus, we have found the Pythagorean triplet (12, 35, 37).

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Verification:

To confirm, we check if the Pythagorean property ($a^2 + b^2 = c^2$) holds true for (12, 35, 37).

$12^2 + 35^2 = 144 + 1225 = 1369$

$37^2 = 1369$

Since $12^2 + 35^2 = 37^2$, our triplet is correct.

We can find the squares using algebraic identities like $(a+b)^2 = a^2 + 2ab + b^2$ and $(a-b)^2 = a^2 - 2ab + b^2$.

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(i) 32

We write $32$ as $(30 + 2)$.

$32^2 = (30 + 2)^2$

Using the identity $(a+b)^2 = a^2 + 2ab + b^2$, where $a=30$ and $b=2$:

$32^2 = 30^2 + 2(30)(2) + 2^2$

$32^2 = 900 + 120 + 4$

$32^2 = 1020 + 4$

$32^2 = \mathbf{1024}$

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(ii) 35

We write $35$ as $(30 + 5)$.

$35^2 = (30 + 5)^2$

Using the identity $(a+b)^2 = a^2 + 2ab + b^2$, where $a=30$ and $b=5$:

$35^2 = 30^2 + 2(30)(5) + 5^2$

$35^2 = 900 + 300 + 25$

$35^2 = 1200 + 25$

$35^2 = \mathbf{1225}$

Alternatively, for numbers ending in 5, like $(n5)$, the square is $n(n+1)$ followed by 25. For 35, $n=3$. $n(n+1) = 3(3+1) = 3 \times 4 = 12$. So, $35^2 = 1225$.

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(iii) 86

We write $86$ as $(80 + 6)$ or $(90 - 4)$. Let's use $(80 + 6)$.

$86^2 = (80 + 6)^2$

Using the identity $(a+b)^2 = a^2 + 2ab + b^2$, where $a=80$ and $b=6$:

$86^2 = 80^2 + 2(80)(6) + 6^2$

$86^2 = 6400 + 960 + 36$

$86^2 = 7360 + 36$

$86^2 = \mathbf{7396}$

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(iv) 93

We write $93$ as $(90 + 3)$ or $(100 - 7)$. Let's use $(90 + 3)$.

$93^2 = (90 + 3)^2$

Using the identity $(a+b)^2 = a^2 + 2ab + b^2$, where $a=90$ and $b=3$:

$93^2 = 90^2 + 2(90)(3) + 3^2$

$93^2 = 8100 + 540 + 9$

$93^2 = 8640 + 9$

$93^2 = \mathbf{8649}$

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(v) 71

We write $71$ as $(70 + 1)$.

$71^2 = (70 + 1)^2$

Using the identity $(a+b)^2 = a^2 + 2ab + b^2$, where $a=70$ and $b=1$:

$71^2 = 70^2 + 2(70)(1) + 1^2$

$71^2 = 4900 + 140 + 1$

$71^2 = 5040 + 1$

$71^2 = \mathbf{5041}$

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(vi) 46

We write $46$ as $(40 + 6)$ or $(50 - 4)$. Let's use $(50 - 4)$.

$46^2 = (50 - 4)^2$

Using the identity $(a-b)^2 = a^2 - 2ab + b^2$, where $a=50$ and $b=4$:

$46^2 = 50^2 - 2(50)(4) + 4^2$

$46^2 = 2500 - 400 + 16$

$46^2 = 2100 + 16$

$46^2 = \mathbf{2116}$

The general form of a Pythagorean triplet is given by $(2m, m^2 - 1, m^2 + 1)$, where $m$ is any natural number greater than 1.

We will find the value of $m$ for each case and then determine the other two members of the triplet.

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(i) One member is 6

Let's assume $2m = 6$.

Then, $m = \frac{6}{2} = 3$.

Since $m=3$ is an integer greater than 1, we can find the other members:

First member = $2m = 6$

Second member = $m^2 - 1 = 3^2 - 1 = 9 - 1 = 8$

Third member = $m^2 + 1 = 3^2 + 1 = 9 + 1 = 10$

Thus, the Pythagorean triplet is (6, 8, 10).

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(ii) One member is 14

Let's assume $2m = 14$.

Then, $m = \frac{14}{2} = 7$.

Since $m=7$ is an integer greater than 1, we can find the other members:

First member = $2m = 14$

Second member = $m^2 - 1 = 7^2 - 1 = 49 - 1 = 48$

Third member = $m^2 + 1 = 7^2 + 1 = 49 + 1 = 50$

Thus, the Pythagorean triplet is (14, 48, 50).

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(iii) One member is 16

Let's assume $2m = 16$.

Then, $m = \frac{16}{2} = 8$.

Since $m=8$ is an integer greater than 1, we can find the other members:

First member = $2m = 16$

Second member = $m^2 - 1 = 8^2 - 1 = 64 - 1 = 63$

Third member = $m^2 + 1 = 8^2 + 1 = 64 + 1 = 65$

Thus, the Pythagorean triplet is (16, 63, 65).

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(iv) One member is 18

Let's assume $2m = 18$.

Then, $m = \frac{18}{2} = 9$.

Since $m=9$ is an integer greater than 1, we can find the other members:

First member = $2m = 18$

Second member = $m^2 - 1 = 9^2 - 1 = 81 - 1 = 80$

Third member = $m^2 + 1 = 9^2 + 1 = 81 + 1 = 82$

Thus, the Pythagorean triplet is (18, 80, 82).

We can find the square root of 6400 using several methods. Here are two common methods:

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Method 1: Prime Factorization

First, we find the prime factorization of 6400.

$$\begin{array}{c|cc} 2 & 6400 \\ \hline 2 & 3200 \\ \hline 2 & 1600 \\ \hline 2 & 800 \\ \hline 2 & 400 \\ \hline 2 & 200 \\ \hline 2 & 100 \\ \hline 2 & 50 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$$

The prime factorization of 6400 is $2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 5 \times 5$.

To find the square root, we group the prime factors into pairs of identical factors:

$6400 = (2 \times 2) \times (2 \times 2) \times (2 \times 2) \times (2 \times 2) \times (5 \times 5)$

For each pair, we take one factor out:

$\sqrt{6400} = 2 \times 2 \times 2 \times 2 \times 5$

$\sqrt{6400} = 16 \times 5$

$\sqrt{6400} = \mathbf{80}$

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Method 2: Using Properties of Square Roots

We can write 6400 as a product of simpler squares.

$6400 = 64 \times 100$

Now, we take the square root:

$\sqrt{6400} = \sqrt{64 \times 100}$

Using the property $\sqrt{a \times b} = \sqrt{a} \times \sqrt{b}$:

$\sqrt{6400} = \sqrt{64} \times \sqrt{100}$

We know that $\sqrt{64} = 8$ (since $8^2 = 64$) and $\sqrt{100} = 10$ (since $10^2 = 100$).

$\sqrt{6400} = 8 \times 10$

$\sqrt{6400} = \mathbf{80}$

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Both methods show that the square root of 6400 is 80.

To determine if 90 is a perfect square, we can find its prime factorization. A number is a perfect square if all of its prime factors occur in pairs.

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Let's find the prime factors of 90:

$$\begin{array}{c|cc} 2 & 90 \\ \hline 3 & 45 \\ \hline 3 & 15 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$$

The prime factorization of 90 is $2 \times 3 \times 3 \times 5$.

We can write this as $90 = 2^1 \times 3^2 \times 5^1$.

To be a perfect square, all the exponents in the prime factorization must be even numbers.

In the factorization of 90, the prime factor 3 has an even exponent (2), but the prime factors 2 and 5 have odd exponents (1).

Since not all prime factors occur in pairs (or have even exponents), 90 is not a perfect square.

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Alternatively, we know that perfect squares only end in the digits 0, 1, 4, 5, 6, or 9. While 90 ends in 0, a number ending in zero is a perfect square only if it ends in an even number of zeros. 90 ends in only one zero (an odd number of zeros). Therefore, 90 is not a perfect square.

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We can also check the squares of integers around the approximate square root of 90. We know $9^2 = 81$ and $10^2 = 100$. Since 90 lies between 81 and 100, there is no integer whose square is 90.

---

Conclusion: No, 90 is not a perfect square.

Step 1: Check if 2352 is a perfect square using prime factorization.

We find the prime factors of 2352:

$$\begin{array}{c|cc} 2 & 2352 \\ \hline 2 & 1176 \\ \hline 2 & 588 \\ \hline 2 & 294 \\ \hline 3 & 147 \\ \hline 7 & 49 \\ \hline 7 & 7 \\ \hline & 1 \end{array}$$

The prime factorization of 2352 is $2 \times 2 \times 2 \times 2 \times 3 \times 7 \times 7 = 2^4 \times 3^1 \times 7^2$.

For a number to be a perfect square, all the exponents in its prime factorization must be even.

In the factorization $2^4 \times 3^1 \times 7^2$, the exponent of the prime factor 3 is 1, which is an odd number.

Therefore, 2352 is not a perfect square.

---

Step 2: Find the smallest multiple of 2352 which is a perfect square.

To make 2352 a perfect square, we need to make the exponents of all its prime factors even.

The prime factorization is $2^4 \times 3^1 \times 7^2$. The only factor with an odd exponent is 3 (exponent 1).

To make the exponent of 3 even, we need to multiply by another factor of $3^1$, which is 3.

The smallest number by which 2352 must be multiplied to become a perfect square is 3.

The new number (the perfect square) is $2352 \times 3$.

New number = $(2^4 \times 3^1 \times 7^2) \times 3 = 2^4 \times 3^2 \times 7^2$.

Calculating the value: $2352 \times 3 = \mathbf{7056}$.

So, the smallest multiple of 2352 that is a perfect square is 7056.

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Step 3: Find the square root of the new number (7056).

The new number is $7056 = 2^4 \times 3^2 \times 7^2$.

To find the square root, we take half of each exponent in the prime factorization:

$\sqrt{7056} = \sqrt{2^4 \times 3^2 \times 7^2}$

$\sqrt{7056} = 2^{(4/2)} \times 3^{(2/2)} \times 7^{(2/2)}$

$\sqrt{7056} = 2^2 \times 3^1 \times 7^1$

$\sqrt{7056} = 4 \times 3 \times 7$

$\sqrt{7056} = 12 \times 7$

$\sqrt{7056} = \mathbf{84}$

---

Summary:

• Is 2352 a perfect square? No.
• The smallest multiple of 2352 which is a perfect square is $2352 \times 3 = \mathbf{7056}$.
• The square root of the new number (7056) is 84.

Step 1: Find the prime factorization of 9408.

We perform prime factorization for 9408:

$$\begin{array}{c|cc} 2 & 9408 \\ \hline 2 & 4704 \\ \hline 2 & 2352 \\ \hline 2 & 1176 \\ \hline 2 & 588 \\ \hline 2 & 294 \\ \hline 3 & 147 \\ \hline 7 & 49 \\ \hline 7 & 7 \\ \hline & 1 \end{array}$$

The prime factorization of 9408 is $2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 7 \times 7$.

So, $9408 = 2^6 \times 3^1 \times 7^2$.

---

Step 2: Identify the prime factors with odd exponents.

For a number to be a perfect square, all the exponents in its prime factorization must be even.

In the factorization $2^6 \times 3^1 \times 7^2$, the exponents are 6, 1, and 2.

The exponent of the prime factor 3 is 1, which is an odd number. The exponents of 2 and 7 are even.

The factor 3 is not paired.

---

Step 3: Find the smallest number by which 9408 must be divided.

To make the quotient a perfect square, we need to remove the factors with odd exponents by division. In this case, we need to divide by the factor $3^1$.

Therefore, the smallest number by which 9408 must be divided to get a perfect square is 3.

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Step 4: Calculate the perfect square quotient.

The quotient is obtained by dividing 9408 by 3:

Quotient = $\frac{9408}{3}$

Quotient = $\frac{2^6 \times 3^1 \times 7^2}{3^1}$

Quotient = $2^6 \times 7^2$

Quotient = $64 \times 49$

Quotient = $\mathbf{3136}$

So, 3136 is the perfect square obtained after dividing 9408 by 3.

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Step 5: Find the square root of the quotient (3136).

The quotient is $3136 = 2^6 \times 7^2$.

To find the square root, we take half of each exponent in the prime factorization:

$\sqrt{3136} = \sqrt{2^6 \times 7^2}$

$\sqrt{3136} = 2^{(6/2)} \times 7^{(2/2)}$

$\sqrt{3136} = 2^3 \times 7^1$

$\sqrt{3136} = 8 \times 7$

$\sqrt{3136} = \mathbf{56}$

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Summary:

• The smallest number by which 9408 must be divided to get a perfect square is 3.
• The resulting perfect square quotient is 3136.
• The square root of the quotient (3136) is 56.

Solution:

The problem requires us to find the smallest square number that is divisible by each of the numbers 6, 9, and 15.

First, we need to find the smallest number which is a multiple of all three numbers. This is their Least Common Multiple (LCM).

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Step 1: Find the LCM of 6, 9, and 15

We will use the division method to find the LCM.

2	6 , 9 , 15
3	3 , 9 , 15
3	1 , 3 , 5
5	1 , 1 , 5
	1 , 1 , 1

The LCM is the product of the divisors:

LCM(6, 9, 15) = $2 \times 3 \times 3 \times 5 = 90$.

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Step 2: Check if the LCM is a perfect square

Now, we examine the prime factorization of the LCM, which is 90.

$90 = 2 \times 3 \times 3 \times 5 = 2^1 \times 3^2 \times 5^1$

For a number to be a perfect square, all of its prime factors must occur in pairs. In the factorization of 90, the prime factor 3 is in a pair ($3^2$), but the prime factors 2 and 5 are single (not in pairs).

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Step 3: Find the smallest square number

To make 90 a perfect square, we must multiply it by the prime factors that are not in pairs. In this case, we need to multiply by another 2 and another 5 to create pairs for them.

So, we multiply 90 by $2 \times 5 = 10$.

The required smallest square number = $90 \times 10 = 900$.

The prime factorization of 900 is $2^2 \times 3^2 \times 5^2$, which confirms it is a perfect square as all factors are in pairs.

Therefore, the smallest square number which is divisible by each of the numbers 6, 9 and 15 is 900.

The unit digit of the square root of a number is determined by the unit digit of the number itself. We need to consider which digits, when squared, produce the given unit digit.

We know the following pattern for the unit digits of squares:

• $0^2$ ends in 0
• $1^2$ ends in 1
• $2^2$ ends in 4
• $3^2$ ends in 9
• $4^2$ ends in 6
• $5^2$ ends in 5
• $6^2$ ends in 6
• $7^2$ ends in 9
• $8^2$ ends in 4
• $9^2$ ends in 1

From this, we can deduce the possible unit digits of the square root:

• If a perfect square ends in 0, its square root ends in 0.
• If a perfect square ends in 1, its square root ends in 1 or 9.
• If a perfect square ends in 4, its square root ends in 2 or 8.
• If a perfect square ends in 5, its square root ends in 5.
• If a perfect square ends in 6, its square root ends in 4 or 6.
• If a perfect square ends in 9, its square root ends in 3 or 7.

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(i) 9801

The unit digit of 9801 is 1.

Therefore, the possible unit digits of its square root are 1 or 9.

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(ii) 99856

The unit digit of 99856 is 6.

Therefore, the possible unit digits of its square root are 4 or 6.

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(iii) 998001

The unit digit of 998001 is 1.

Therefore, the possible unit digits of its square root are 1 or 9.

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(iv) 657666025

The unit digit of 657666025 is 5.

Therefore, the possible unit digit of its square root is 5.

Reasoning:

A fundamental property of perfect squares is related to their unit digit (the digit in the one's place). A number that is a perfect square can only end in the digits 0, 1, 4, 5, 6, or 9.

Conversely, any integer ending in 2, 3, 7, or 8 is surely not a perfect square.

We will examine the unit digit of each given number to determine if it can possibly be a perfect square.

---

(i) 153

The unit digit of 153 is 3.

Since numbers ending in 3 cannot be perfect squares, 153 is surely not a perfect square.

---

(ii) 257

The unit digit of 257 is 7.

Since numbers ending in 7 cannot be perfect squares, 257 is surely not a perfect square.

---

(iii) 408

The unit digit of 408 is 8.

Since numbers ending in 8 cannot be perfect squares, 408 is surely not a perfect square.

---

(iv) 441

The unit digit of 441 is 1.

Numbers ending in 1 can be perfect squares (e.g., $1^2=1$, $9^2=81$, $11^2=121$, $21^2=441$).

Therefore, based solely on the unit digit, we cannot be sure that 441 is not a perfect square. (In fact, 441 is a perfect square, $21^2=441$).

---

Conclusion:

The numbers which are surely not perfect squares based on their unit digits are 153, 257, and 408.

The method of repeated subtraction relies on the property that every perfect square $n^2$ is the sum of the first $n$ consecutive odd natural numbers. Therefore, if we subtract consecutive odd numbers (1, 3, 5, ...) from a perfect square, we will reach zero. The number of steps (subtractions) taken will be the square root of the number.

---

(i) Finding the square root of 100

We subtract consecutive odd numbers starting from 1:

1. $100 - 1 = 99$

2. $99 - 3 = 96$

3. $96 - 5 = 91$

4. $91 - 7 = 84$

5. $84 - 9 = 75$

6. $75 - 11 = 64$

7. $64 - 13 = 51$

8. $51 - 15 = 36$

9. $36 - 17 = 19$

10. $19 - 19 = 0$

We reached 0 after 10 steps (subtractions).

Therefore, the square root of 100 is 10.

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(ii) Finding the square root of 169

We subtract consecutive odd numbers starting from 1:

1. $169 - 1 = 168$

2. $168 - 3 = 165$

3. $165 - 5 = 160$

4. $160 - 7 = 153$

5. $153 - 9 = 144$

6. $144 - 11 = 133$

7. $133 - 13 = 120$

8. $120 - 15 = 105$

9. $105 - 17 = 88$

10. $88 - 19 = 69$

11. $69 - 21 = 48$

12. $48 - 23 = 25$

13. $25 - 25 = 0$

We reached 0 after 13 steps (subtractions).

Therefore, the square root of 169 is 13.

The Prime Factorization Method involves finding the prime factors of the number, grouping them into pairs of equal factors, and then taking one factor from each pair and multiplying them to get the square root.

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(i) 729

Prime factorization of 729:

$\begin{array}{c|cc} 3 & 729 \\ \hline 3 & 243 \\ \hline 3 & 81 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

$729 = 3 \times 3 \times 3 \times 3 \times 3 \times 3 = (3 \times 3) \times (3 \times 3) \times (3 \times 3)$

Taking one factor from each pair: $3 \times 3 \times 3 = 27$.

Therefore, $\sqrt{729} = \mathbf{27}$.

---

(ii) 400

Prime factorization of 400:

$\begin{array}{c|cc} 2 & 400 \\ \hline 2 & 200 \\ \hline 2 & 100 \\ \hline 2 & 50 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

$400 = 2 \times 2 \times 2 \times 2 \times 5 \times 5 = (2 \times 2) \times (2 \times 2) \times (5 \times 5)$

Taking one factor from each pair: $2 \times 2 \times 5 = 20$.

Therefore, $\sqrt{400} = \mathbf{20}$.

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(iii) 1764

Prime factorization of 1764:

$\begin{array}{c|cc} 2 & 1764 \\ \hline 2 & 882 \\ \hline 3 & 441 \\ \hline 3 & 147 \\ \hline 7 & 49 \\ \hline 7 & 7 \\ \hline & 1 \end{array}$

$1764 = 2 \times 2 \times 3 \times 3 \times 7 \times 7 = (2 \times 2) \times (3 \times 3) \times (7 \times 7)$

Taking one factor from each pair: $2 \times 3 \times 7 = 42$.

Therefore, $\sqrt{1764} = \mathbf{42}$.

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(iv) 4096

Prime factorization of 4096:

$\begin{array}{c|cc} 2 & 4096 \\ \hline 2 & 2048 \\ \hline 2 & 1024 \\ \hline 2 & 512 \\ \hline 2 & 256 \\ \hline 2 & 128 \\ \hline 2 & 64 \\ \hline 2 & 32 \\ \hline 2 & 16 \\ \hline 2 & 8 \\ \hline 2 & 4 \\ \hline 2 & 2 \\ \hline & 1 \end{array}$

$4096 = 2^{12} = (2 \times 2) \times (2 \times 2) \times (2 \times 2) \times (2 \times 2) \times (2 \times 2) \times (2 \times 2)$

Taking one factor from each pair: $2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^6 = 64$.

Therefore, $\sqrt{4096} = \mathbf{64}$.

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(v) 7744

Prime factorization of 7744:

$\begin{array}{c|cc} 2 & 7744 \\ \hline 2 & 3872 \\ \hline 2 & 1936 \\ \hline 2 & 968 \\ \hline 2 & 484 \\ \hline 2 & 242 \\ \hline 11 & 121 \\ \hline 11 & 11 \\ \hline & 1 \end{array}$

$7744 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 11 \times 11 \ $$ = (2 \times 2) \times (2 \times 2) \times (2 \times 2) \times (11 \times 11)$

Taking one factor from each pair: $2 \times 2 \times 2 \times 11 = 8 \times 11 = 88$.

Therefore, $\sqrt{7744} = \mathbf{88}$.

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(vi) 9604

Prime factorization of 9604:

$\begin{array}{c|cc} 2 & 9604 \\ \hline 2 & 4802 \\ \hline 7 & 2401 \\ \hline 7 & 343 \\ \hline 7 & 49 \\ \hline 7 & 7 \\ \hline & 1 \end{array}$

$9604 = 2 \times 2 \times 7 \times 7 \times 7 \times 7 = (2 \times 2) \times (7 \times 7) \times (7 \times 7)$

Taking one factor from each pair: $2 \times 7 \times 7 = 2 \times 49 = 98$.

Therefore, $\sqrt{9604} = \mathbf{98}$.

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(vii) 5929

Prime factorization of 5929:

$\begin{array}{c|cc} 7 & 5929 \\ \hline 7 & 847 \\ \hline 11 & 121 \\ \hline 11 & 11 \\ \hline & 1 \end{array}$

$5929 = 7 \times 7 \times 11 \times 11 = (7 \times 7) \times (11 \times 11)$

Taking one factor from each pair: $7 \times 11 = 77$.

Therefore, $\sqrt{5929} = \mathbf{77}$.

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(viii) 9216

Prime factorization of 9216:

$\begin{array}{c|cc} 2 & 9216 \\ \hline 2 & 4608 \\ \hline 2 & 2304 \\ \hline 2 & 1152 \\ \hline 2 & 576 \\ \hline 2 & 288 \\ \hline 2 & 144 \\ \hline 2 & 72 \\ \hline 2 & 36 \\ \hline 2 & 18 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

$9216 = 2^{10} \times 3^2 \ $$ = (2 \times 2) \times (2 \times 2) \times (2 \times 2) \times (2 \times 2) \times (2 \times 2) \times (3 \times 3)$

Taking one factor from each pair: $2 \times 2 \times 2 \times 2 \times 2 \times 3 = 2^5 \times 3 = 32 \times 3 = 96$.

Therefore, $\sqrt{9216} = \mathbf{96}$.

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(ix) 529

Prime factorization of 529:

$\begin{array}{c|cc} 23 & 529 \\ \hline 23 & 23 \\ \hline & 1 \end{array}$

$529 = 23 \times 23 = (23 \times 23)$

Taking one factor from each pair: $23$.

Therefore, $\sqrt{529} = \mathbf{23}$.

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(x) 8100

Prime factorization of 8100:

$\begin{array}{c|cc} 2 & 8100 \\ \hline 2 & 4050 \\ \hline 3 & 2025 \\ \hline 3 & 675 \\ \hline 3 & 225 \\ \hline 3 & 75 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

$8100 = 2 \times 2 \times 3 \times 3 \times 3 \times 3 \times 5 \times 5 \ $$ = (2 \times 2) \times (3 \times 3) \times (3 \times 3) \times (5 \times 5)$

Taking one factor from each pair: $2 \times 3 \times 3 \times 5 = 2 \times 9 \times 5 = 18 \times 5 = 90$.

Therefore, $\sqrt{8100} = \mathbf{90}$.

To solve this, we first find the prime factorization of each number. If any prime factor appears an odd number of times, the number is not a perfect square. To make it a perfect square, we multiply by the product of the prime factors that have odd exponents. The smallest whole number to multiply by is this product. Then we find the square root of the resulting perfect square.

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(i) 252

1. Prime Factorization:

$\begin{array}{c|cc} 2 & 252 \\ \hline 2 & 126 \\ \hline 3 & 63 \\ \hline 3 & 21 \\ \hline 7 & 7 \\ \hline & 1 \end{array}$

$252 = 2 \times 2 \times 3 \times 3 \times 7 = 2^2 \times 3^2 \times 7^1$

2. Smallest Multiplier: The prime factor 7 has an odd exponent (1). To make it a perfect square, we need to multiply by 7.

Smallest whole number to multiply by = 7.

3. Perfect Square Obtained:

$252 \times 7 = (2^2 \times 3^2 \times 7^1) \times 7 = 2^2 \times 3^2 \times 7^2$

$252 \times 7 = 1764$. The perfect square is 1764.

4. Square Root of the New Number:

$\sqrt{1764} = \sqrt{2^2 \times 3^2 \times 7^2} = 2^1 \times 3^1 \times 7^1 = 2 \times 3 \times 7 = \mathbf{42}$.

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(ii) 180

1. Prime Factorization:

$\begin{array}{c|cc} 2 & 180 \\ \hline 2 & 90 \\ \hline 3 & 45 \\ \hline 3 & 15 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

$180 = 2 \times 2 \times 3 \times 3 \times 5 = 2^2 \times 3^2 \times 5^1$

2. Smallest Multiplier: The prime factor 5 has an odd exponent (1). Multiply by 5.

Smallest whole number to multiply by = 5.

3. Perfect Square Obtained:

$180 \times 5 = (2^2 \times 3^2 \times 5^1) \times 5 = 2^2 \times 3^2 \times 5^2$

$180 \times 5 = 900$. The perfect square is 900.

4. Square Root of the New Number:

$\sqrt{900} = \sqrt{2^2 \times 3^2 \times 5^2} = 2^1 \times 3^1 \times 5^1 = 2 \times 3 \times 5 = \mathbf{30}$.

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(iii) 1008

1. Prime Factorization:

$\begin{array}{c|cc} 2 & 1008 \\ \hline 2 & 504 \\ \hline 2 & 252 \\ \hline 2 & 126 \\ \hline 3 & 63 \\ \hline 3 & 21 \\ \hline 7 & 7 \\ \hline & 1 \end{array}$

$1008 = 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 7 = 2^4 \times 3^2 \times 7^1$

2. Smallest Multiplier: The prime factor 7 has an odd exponent (1). Multiply by 7.

Smallest whole number to multiply by = 7.

3. Perfect Square Obtained:

$1008 \times 7 = (2^4 \times 3^2 \times 7^1) \times 7 = 2^4 \times 3^2 \times 7^2$

$1008 \times 7 = 7056$. The perfect square is 7056.

4. Square Root of the New Number:

$\sqrt{7056} = \sqrt{2^4 \times 3^2 \times 7^2} = 2^2 \times 3^1 \times 7^1 = 4 \times 3 \times 7 = \mathbf{84}$.

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(iv) 2028

1. Prime Factorization:

$\begin{array}{c|cc} 2 & 2028 \\ \hline 2 & 1014 \\ \hline 3 & 507 \\ \hline 13 & 169 \\ \hline 13 & 13 \\ \hline & 1 \end{array}$

$2028 = 2 \times 2 \times 3 \times 13 \times 13 = 2^2 \times 3^1 \times 13^2$

2. Smallest Multiplier: The prime factor 3 has an odd exponent (1). Multiply by 3.

Smallest whole number to multiply by = 3.

3. Perfect Square Obtained:

$2028 \times 3 = (2^2 \times 3^1 \times 13^2) \times 3 = 2^2 \times 3^2 \times 13^2$

$2028 \times 3 = 6084$. The perfect square is 6084.

4. Square Root of the New Number:

$\sqrt{6084} = \sqrt{2^2 \times 3^2 \times 13^2} = 2^1 \times 3^1 \times 13^1 = 2 \times 3 \times 13 = \mathbf{78}$.

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(v) 1458

1. Prime Factorization:

$\begin{array}{c|cc} 2 & 1458 \\ \hline 3 & 729 \\ \hline 3 & 243 \\ \hline 3 & 81 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

$1458 = 2 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3 = 2^1 \times 3^6$

2. Smallest Multiplier: The prime factor 2 has an odd exponent (1). Multiply by 2.

Smallest whole number to multiply by = 2.

3. Perfect Square Obtained:

$1458 \times 2 = (2^1 \times 3^6) \times 2 = 2^2 \times 3^6$

$1458 \times 2 = 2916$. The perfect square is 2916.

4. Square Root of the New Number:

$\sqrt{2916} = \sqrt{2^2 \times 3^6} = 2^1 \times 3^3 = 2 \times 27 = \mathbf{54}$.

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(vi) 768

1. Prime Factorization:

$\begin{array}{c|cc} 2 & 768 \\ \hline 2 & 384 \\ \hline 2 & 192 \\ \hline 2 & 96 \\ \hline 2 & 48 \\ \hline 2 & 24 \\ \hline 2 & 12 \\ \hline 2 & 6 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

$768 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 = 2^8 \times 3^1$

2. Smallest Multiplier: The prime factor 3 has an odd exponent (1). Multiply by 3.

Smallest whole number to multiply by = 3.

3. Perfect Square Obtained:

$768 \times 3 = (2^8 \times 3^1) \times 3 = 2^8 \times 3^2$

$768 \times 3 = 2304$. The perfect square is 2304.

4. Square Root of the New Number:

$\sqrt{2304} = \sqrt{2^8 \times 3^2} = 2^4 \times 3^1 = 16 \times 3 = \mathbf{48}$.

The process involves finding the prime factorization of the number. Identify the prime factors that appear an odd number of times (i.e., have odd exponents). The smallest whole number to divide by is the product of these unpaired factors. Dividing the original number by this smallest whole number yields a perfect square quotient. We then find the square root of this quotient.

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(i) 252

1. Prime Factorization:

$\begin{array}{c|cc} 2 & 252 \\ \hline 2 & 126 \\ \hline 3 & 63 \\ \hline 3 & 21 \\ \hline 7 & 7 \\ \hline & 1 \end{array}$

$252 = 2 \times 2 \times 3 \times 3 \times 7 = 2^2 \times 3^2 \times 7^1$

2. Smallest Divisor: The prime factor 7 has an odd exponent (1). Thus, the smallest whole number to divide by is 7.

3. Perfect Square Quotient:

$252 \div 7 = \frac{2^2 \times 3^2 \times 7^1}{7^1} = 2^2 \times 3^2 = 4 \times 9 = 36$. The quotient is 36.

4. Square Root of the Quotient:

$\sqrt{36} = \sqrt{2^2 \times 3^2} = 2 \times 3 = \mathbf{6}$.

---

(ii) 2925

1. Prime Factorization:

$\begin{array}{c|cc} 3 & 2925 \\ \hline 3 & 975 \\ \hline 5 & 325 \\ \hline 5 & 65 \\ \hline 13 & 13 \\ \hline & 1 \end{array}$

$2925 = 3 \times 3 \times 5 \times 5 \times 13 = 3^2 \times 5^2 \times 13^1$

2. Smallest Divisor: The prime factor 13 has an odd exponent (1). The smallest whole number to divide by is 13.

3. Perfect Square Quotient:

$2925 \div 13 = \frac{3^2 \times 5^2 \times 13^1}{13^1} = 3^2 \times 5^2 = 9 \times 25 = 225$. The quotient is 225.

4. Square Root of the Quotient:

$\sqrt{225} = \sqrt{3^2 \times 5^2} = 3 \times 5 = \mathbf{15}$.

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(iii) 396

1. Prime Factorization:

$\begin{array}{c|cc} 2 & 396 \\ \hline 2 & 198 \\ \hline 3 & 99 \\ \hline 3 & 33 \\ \hline 11 & 11 \\ \hline & 1 \end{array}$

$396 = 2 \times 2 \times 3 \times 3 \times 11 = 2^2 \times 3^2 \times 11^1$

2. Smallest Divisor: The prime factor 11 has an odd exponent (1). The smallest whole number to divide by is 11.

3. Perfect Square Quotient:

$396 \div 11 = \frac{2^2 \times 3^2 \times 11^1}{11^1} = 2^2 \times 3^2 = 4 \times 9 = 36$. The quotient is 36.

4. Square Root of the Quotient:

$\sqrt{36} = \sqrt{2^2 \times 3^2} = 2 \times 3 = \mathbf{6}$.

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(iv) 2645

1. Prime Factorization:

$\begin{array}{c|cc} 5 & 2645 \\ \hline 23 & 529 \\ \hline 23 & 23 \\ \hline & 1 \end{array}$

$2645 = 5 \times 23 \times 23 = 5^1 \times 23^2$

2. Smallest Divisor: The prime factor 5 has an odd exponent (1). The smallest whole number to divide by is 5.

3. Perfect Square Quotient:

$2645 \div 5 = \frac{5^1 \times 23^2}{5^1} = 23^2 = 529$. The quotient is 529.

4. Square Root of the Quotient:

$\sqrt{529} = \sqrt{23^2} = \mathbf{23}$.

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(v) 2800

1. Prime Factorization:

$\begin{array}{c|cc} 2 & 2800 \\ \hline 2 & 1400 \\ \hline 2 & 700 \\ \hline 2 & 350 \\ \hline 5 & 175 \\ \hline 5 & 35 \\ \hline 7 & 7 \\ \hline & 1 \end{array}$

$2800 = 2 \times 2 \times 2 \times 2 \times 5 \times 5 \times 7 = 2^4 \times 5^2 \times 7^1$

2. Smallest Divisor: The prime factor 7 has an odd exponent (1). The smallest whole number to divide by is 7.

3. Perfect Square Quotient:

$2800 \div 7 = \frac{2^4 \times 5^2 \times 7^1}{7^1} = 2^4 \times 5^2 = 16 \times 25 = 400$. The quotient is 400.

4. Square Root of the Quotient:

$\sqrt{400} = \sqrt{2^4 \times 5^2} = 2^2 \times 5^1 = 4 \times 5 = \mathbf{20}$.

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(vi) 1620

1. Prime Factorization:

$\begin{array}{c|cc} 2 & 1620 \\ \hline 2 & 810 \\ \hline 3 & 405 \\ \hline 3 & 135 \\ \hline 3 & 45 \\ \hline 3 & 15 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

$1620 = 2 \times 2 \times 3 \times 3 \times 3 \times 3 \times 5 = 2^2 \times 3^4 \times 5^1$

2. Smallest Divisor: The prime factor 5 has an odd exponent (1). The smallest whole number to divide by is 5.

3. Perfect Square Quotient:

$1620 \div 5 = \frac{2^2 \times 3^4 \times 5^1}{5^1} = 2^2 \times 3^4 = 4 \times 81 = 324$. The quotient is 324.

4. Square Root of the Quotient:

$\sqrt{324} = \sqrt{2^2 \times 3^4} = 2^1 \times 3^2 = 2 \times 9 = \mathbf{18}$.

Given:

Total amount donated = $\textsf{₹ } 2401$.

Each student donated as many rupees as the total number of students in the class.

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To Find:

The number of students in the class.

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Solution:

Let the number of students in the class be $x$.

According to the problem statement, the amount donated by each student is equal to the number of students, which is $\textsf{₹ } x$.

The total amount donated is the product of the number of students and the amount donated by each student.

Total Donation = (Number of students) $\times$ (Amount donated by each student)

Total Donation = $x \times x = x^2$.

We are given that the total donation is $\textsf{₹ } 2401$.

Therefore, $x^2 = 2401$.

To find the number of students, $x$, we need to calculate the square root of 2401.

$x = \sqrt{2401}$

We can find the square root using the prime factorization method:

$\begin{array}{c|cc} 7 & 2401 \\ \hline 7 & 343 \\ \hline 7 & 49 \\ \hline 7 & 7 \\ \hline & 1 \end{array}$

The prime factorization of 2401 is $7 \times 7 \times 7 \times 7$.

We group the factors into pairs:

$2401 = (7 \times 7) \times (7 \times 7)$

Taking one factor from each pair to find the square root:

$x = \sqrt{2401} = 7 \times 7$

$x = 49$

Alternatively, using the long division method:

$\begin{array}{c|cc} & 4 \ 9 & \\ \hline 4 & \overline{24} \; \overline{01} \\ + \; 4 & 16\phantom{()} \\ \hline 8 \; 9 & 8 \; 01 \\ + \; 9 & 8 \; 01 \\ \hline 98 & 0 \end{array}$

The square root of 2401 is 49.

So, $x = 49$.

This means the number of students in the class is 49.

---

Final Answer:

The number of students in the class is 49.

Given:

Total number of plants = 2025.

Arrangement: The number of plants in each row is equal to the number of rows.

---

To Find:

The number of rows.

The number of plants in each row.

---

Solution:

Let the number of rows in the garden be denoted by $x$.

According to the problem statement, the number of plants in each row is also equal to the number of rows, so it is also $x$.

The total number of plants is calculated by multiplying the number of rows by the number of plants in each row.

Total plants = (Number of rows) $\times$ (Number of plants in each row)

Total plants = $x \times x$

$2025 = x^2$

To find the value of $x$, we need to calculate the square root of 2025.

$x = \sqrt{2025}$

We will use the prime factorization method to find the square root of 2025.

$\begin{array}{c|cc} 3 & 2025 \\ \hline 3 & 675 \\ \hline 3 & 225 \\ \hline 3 & 75 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

The prime factorization of 2025 is $3 \times 3 \times 3 \times 3 \times 5 \times 5$.

We can group the factors into pairs:

$2025 = (3 \times 3) \times (3 \times 3) \times (5 \times 5)$

To find the square root, we take one factor from each pair and multiply them:

$x = \sqrt{2025} = 3 \times 3 \times 5$

$x = 9 \times 5$

$x = 45$

Thus, the number of rows is 45.

Since the number of plants in each row is equal to the number of rows, the number of plants in each row is also 45.

---

Final Answer:

The number of rows is 45.

The number of plants in each row is 45.

To Find:

The smallest number that is a perfect square and is also exactly divisible by 4, 9, and 10.

---

Solution:

Step 1: Find the Least Common Multiple (LCM) of 4, 9, and 10.

Any number that is divisible by 4, 9, and 10 must be a multiple of their LCM. The smallest such number is the LCM itself.

First, we find the prime factorization of each number:

$4 = 2 \times 2 = 2^2$

$9 = 3 \times 3 = 3^2$

$10 = 2 \times 5 = 2^1 \times 5^1$

To find the LCM, we take the highest power of each prime factor that appears in any of the factorizations (the prime factors are 2, 3, and 5).

Highest power of 2 is $2^2$.

Highest power of 3 is $3^2$.

Highest power of 5 is $5^1$.

LCM(4, 9, 10) = $2^2 \times 3^2 \times 5^1$

LCM(4, 9, 10) = $4 \times 9 \times 5$

LCM(4, 9, 10) = $36 \times 5$

LCM(4, 9, 10) = 180.

So, the smallest number divisible by 4, 9, and 10 is 180.

---

Step 2: Determine the smallest square number that is a multiple of the LCM (180).

The number we are looking for must be a multiple of 180 and also a perfect square.

We examine the prime factorization of the LCM: $180 = 2^2 \times 3^2 \times 5^1$.

A number is a perfect square if all the exponents in its prime factorization are even.

In the factorization of 180, the exponents of 2 and 3 are 2 (even), but the exponent of 5 is 1 (odd).

Therefore, 180 is not a perfect square.

To make 180 a perfect square, we need to multiply it by the prime factors that have odd exponents, to make their exponents even. In this case, we need to multiply by $5^1$ (or 5).

The smallest square number that is a multiple of 180 is $180 \times 5$.

Smallest square number = $(2^2 \times 3^2 \times 5^1) \times 5^1$

Smallest square number = $2^2 \times 3^2 \times 5^2$

Smallest square number = $4 \times 9 \times 25$

Smallest square number = $36 \times 25$

Smallest square number = 900.

The number 900 is a perfect square ($30^2 = 900$). It is also divisible by 4 ($900 \div 4 = 225$), 9 ($900 \div 9 = 100$), and 10 ($900 \div 10 = 90$).

---

Final Answer:

The smallest square number that is divisible by each of the numbers 4, 9 and 10 is 900.

To Find:

The smallest number that is a perfect square and is exactly divisible by 8, 15, and 20.

---

Solution:

Step 1: Find the Least Common Multiple (LCM) of 8, 15, and 20.

The smallest number divisible by 8, 15, and 20 is their LCM. Any number divisible by these three must be a multiple of their LCM.

First, find the prime factorization of each number:

$8 = 2 \times 2 \times 2 = 2^3$

$15 = 3 \times 5 = 3^1 \times 5^1$

$20 = 2 \times 2 \times 5 = 2^2 \times 5^1$

To find the LCM, we take the highest power of each prime factor present in the numbers (the prime factors are 2, 3, and 5).

Highest power of 2 is $2^3$.

Highest power of 3 is $3^1$.

Highest power of 5 is $5^1$.

LCM(8, 15, 20) = $2^3 \times 3^1 \times 5^1$

LCM(8, 15, 20) = $8 \times 3 \times 5$

LCM(8, 15, 20) = $24 \times 5$

LCM(8, 15, 20) = 120.

So, the smallest number divisible by 8, 15, and 20 is 120.

---

Step 2: Determine the smallest square number that is a multiple of the LCM (120).

The required number must be a multiple of 120 and also a perfect square.

We examine the prime factorization of the LCM: $120 = 2^3 \times 3^1 \times 5^1$.

For a number to be a perfect square, all the exponents in its prime factorization must be even.

In the factorization of 120, the exponents of 2, 3, and 5 are 3, 1, and 1, respectively. All these exponents are odd.

Therefore, 120 is not a perfect square.

To make 120 a perfect square, we need to multiply it by the prime factors that have odd exponents, so that their exponents become even. We need to multiply by:

• $2^1$ (to make the exponent of 2 become $3+1=4$, which is even)
• $3^1$ (to make the exponent of 3 become $1+1=2$, which is even)
• $5^1$ (to make the exponent of 5 become $1+1=2$, which is even)

The smallest number we need to multiply 120 by is $2^1 \times 3^1 \times 5^1 = 2 \times 3 \times 5 = 30$.

The smallest square number that is a multiple of 120 is $120 \times 30$.

Smallest square number = $(2^3 \times 3^1 \times 5^1) \times (2^1 \times 3^1 \times 5^1)$

Smallest square number = $2^{3+1} \times 3^{1+1} \times 5^{1+1}$

Smallest square number = $2^4 \times 3^2 \times 5^2$

Smallest square number = $16 \times 9 \times 25$

Smallest square number = $144 \times 25$

Smallest square number = 3600.

The number 3600 is a perfect square ($60^2 = 3600$). It is also divisible by 8 ($3600 \div 8 = 450$), 15 ($3600 \div 15 = 240$), and 20 ($3600 \div 20 = 180$).

---

Final Answer:

The smallest square number that is divisible by each of the numbers 8, 15 and 20 is 3600.

We will find the square root of the given numbers using the prime factorization method and the long division method.

---

Method 1: Prime Factorization

---

(i) Finding the square root of 729

First, find the prime factors of 729:

$\begin{array}{c|cc} 3 & 729 \\ \hline 3 & 243 \\ \hline 3 & 81 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

So, the prime factorization of 729 is $3 \times 3 \times 3 \times 3 \times 3 \times 3$.

To find the square root, we group the factors into pairs of equal numbers:

$729 = (3 \times 3) \times (3 \times 3) \times (3 \times 3)$

Taking one factor from each pair:

$\sqrt{729} = 3 \times 3 \times 3$

$\sqrt{729} = 27$

Therefore, the square root of 729 is 27.

---

(ii) Finding the square root of 1296

First, find the prime factors of 1296:

$\begin{array}{c|cc} 2 & 1296 \\ \hline 2 & 648 \\ \hline 2 & 324 \\ \hline 2 & 162 \\ \hline 3 & 81 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

So, the prime factorization of 1296 is $2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3$.

To find the square root, we group the factors into pairs of equal numbers:

$1296 = (2 \times 2) \times (2 \times 2) \times (3 \times 3) \times (3 \times 3)$

Taking one factor from each pair:

$\sqrt{1296} = 2 \times 2 \times 3 \times 3$

$\sqrt{1296} = 4 \times 9$

$\sqrt{1296} = 36$

Therefore, the square root of 1296 is 36.

---

Method 2: Long Division

---

(i) Finding the square root of 729

We use the long division method as follows:

$\begin{array}{c|cc} & 2 \ 7 & \\ \hline 2 & \overline{7} \; \overline{29} \\ + \; 2 & 4\phantom{()} \\ \hline 4 \; 7 & 3 \; 29 \\ + \; 7 & 3 \; 29 \\ \hline 54 & 0 \end{array}$

Steps:

1. Group the digits in pairs from right to left: $\overline{7} \; \overline{29}$.

2. Find the largest number whose square is less than or equal to the first group (7). This is 2 ($2^2=4$). Write 2 as the divisor and the first digit of the quotient.

3. Subtract the square ($4$) from the first group ($7-4=3$).

4. Bring down the next pair (29) to form the new dividend 329.

5. Double the quotient (2) to get 4. Find the largest digit $y$ such that $4y \times y \le 329$. We try $y=7$. $47 \times 7 = 329$.

6. Write 7 as the next digit of the quotient and also add it to the divisor. The new divisor is 47.

7. Multiply the new divisor (47) by the new digit in the quotient (7): $47 \times 7 = 329$. Subtract this from the dividend (329). The remainder is 0.

Since the remainder is 0, the square root of 729 is 27.

---

(ii) Finding the square root of 1296

We use the long division method as follows:

$\begin{array}{c|cc} & 3 \ 6 & \\ \hline 3 & \overline{12} \; \overline{96} \\ + \; 3 & 9\phantom{()} \\ \hline 6 \; 6 & 3 \; 96 \\ + \; 6 & 3 \; 96 \\ \hline 72 & 0 \end{array}$

Steps:

1. Group the digits in pairs from right to left: $\overline{12} \; \overline{96}$.

2. Find the largest number whose square is less than or equal to the first group (12). This is 3 ($3^2=9$). Write 3 as the divisor and the first digit of the quotient.

3. Subtract the square ($9$) from the first group ($12-9=3$).

4. Bring down the next pair (96) to form the new dividend 396.

5. Double the quotient (3) to get 6. Find the largest digit $y$ such that $6y \times y \le 396$. We try $y=6$. $66 \times 6 = 396$.

6. Write 6 as the next digit of the quotient and also add it to the divisor. The new divisor is 66.

7. Multiply the new divisor (66) by the new digit in the quotient (6): $66 \times 6 = 396$. Subtract this from the dividend (396). The remainder is 0.

Since the remainder is 0, the square root of 1296 is 36.

To Find:

1. The least number that must be subtracted from 5607 to get a perfect square.

2. The square root of the resulting perfect square.

---

Solution:

To find the least number that must be subtracted from 5607 to make it a perfect square, we first need to find the square root of 5607 using the long division method.

We perform the long division as follows:

$\begin{array}{c|cc} & 7 \ 4 & \\ \hline 7 & \overline{56} \; \overline{07} \\ + \; 7 & 49\phantom{()} \\ \hline 14 \; 4 & 7 \; 07 \\ + \; 4 & 5 \; 76 \\ \hline 148 & 1 \; 31 \\ \end{array}$

Steps of Long Division:

1. Pair the digits of 5607 starting from the unit's place (right to left). We get two pairs: $\overline{56}$ and $\overline{07}$.

2. Find the largest number whose square is less than or equal to the first pair (56). The number is 7, as $7^2 = 49$. Write 7 in the quotient's place and also as the divisor.

3. Subtract the square ($49$) from the first pair ($56 - 49 = 7$).

4. Bring down the next pair (07) to the right of the remainder (7) to get the new dividend, which is 707.

5. Double the current quotient (7) to get $2 \times 7 = 14$. Place this number as the first part of the new divisor.

6. Find the largest digit $y$ such that the product of the new divisor (formed by placing $y$ next to 14, i.e., $14y$) and $y$ is less than or equal to the new dividend (707).

If we try $y=4$, we get $144 \times 4 = 576$.

If we try $y=5$, we get $145 \times 5 = 725$. Since $725 > 707$, the required digit is 4.

7. Write 4 in the quotient next to 7, making the quotient 74. Write 4 next to 14 in the divisor place, making the new divisor 144.

8. Multiply the new divisor (144) by the new digit in the quotient (4): $144 \times 4 = 576$. Subtract this product from the dividend ($707 - 576 = 131$).

The remainder obtained is 131.

This remainder (131) is the least number that must be subtracted from 5607 to make the result a perfect square.

The perfect square number is $5607 - 131 = 5476$.

The square root of this perfect square is the quotient obtained in the long division process, which is 74.

We can check this: $74^2 = 74 \times 74 = 5476$.

---

Final Answer:

The least number that must be subtracted from 5607 to get a perfect square is 131.

The perfect square obtained is $5607 - 131 = \mathbf{5476}$.

The square root of the perfect square (5476) is 74.

To Find:

The largest 4-digit number that is a perfect square.

---

Solution:

The greatest 4-digit number is 9999.

To find the greatest 4-digit perfect square, we first find the square root of 9999 using the long division method. The integer part of this square root will give us the largest integer whose square is less than or equal to 9999.

Using the long division method for finding the square root of 9999:

$\begin{array}{c|cc} & 9 \ 9 & \\ \hline 9 & \overline{99} \; \overline{99} \\ + \; 9 & 81\phantom{()} \\ \hline 18 \; 9 & 18 \; 99 \\ + \; 9 & 17 \; 01 \\ \hline 198 & \phantom{0}1 \; 98 \\ \end{array}$

From the long division, we see that $\sqrt{9999}$ is slightly more than 99, and the remainder is 198.

This means that $99^2$ is the largest perfect square less than 9999.

Alternatively, we can subtract the remainder (198) from 9999 to get the largest perfect square less than 9999.

Required perfect square = $9999 - 198 = 9801$.

Let's calculate $99^2$ to verify:

$99^2 = 99 \times 99$

We can use the identity $(a-b)^2 = a^2 - 2ab + b^2$:

$99^2 = (100 - 1)^2 = 100^2 - 2(100)(1) + 1^2$

$99^2 = 10000 - 200 + 1$

$99^2 = 9800 + 1$

$99^2 = 9801$.

The next perfect square is $100^2 = 10000$, which is a 5-digit number.

Therefore, 9801 is the greatest 4-digit number which is a perfect square.

---

Final Answer:

The greatest 4-digit number which is a perfect square is 9801.

To Find:

1. The least number that must be added to 1300 to get a perfect square.

2. The square root of the resulting perfect square.

---

Solution:

We need to find the smallest perfect square number that is greater than 1300.

First, we find the approximate square root of 1300 using the long division method.

$\begin{array}{c|cc} & 3 \ 6 & \\ \hline 3 & \overline{13} \; \overline{00} \\ + \; 3 & 9\phantom{()} \\ \hline 6 \; 6 & 4 \; 00 \\ + \; 6 & 3 \; 96 \\ \hline 72 & \phantom{00}4 \\ \end{array}$

The long division shows that $36^2$ is less than 1300, as there is a non-zero remainder (4).

Specifically, $36^2 = 1296$.

Since $36^2 = 1296 < 1300$, the smallest perfect square greater than 1300 will be the square of the next integer, which is $(36 + 1) = 37$.

Let's calculate the square of 37:

$37^2 = 37 \times 37$

$37^2 = 1369$.

So, the smallest perfect square greater than 1300 is 1369.

Now, we find the number that must be added to 1300 to get this perfect square (1369).

Number to be added = (Next perfect square) - (Given number)

Number to be added = $1369 - 1300$

Number to be added = 69.

The perfect square obtained by adding 69 to 1300 is $1300 + 69 = \mathbf{1369}$.

The square root of this perfect square is the integer whose square we calculated, which is 37.

$\sqrt{1369} = 37$.

---

Final Answer:

The least number that must be added to 1300 to get a perfect square is 69.

The perfect square obtained is 1369.

The square root of the perfect square (1369) is 37.

To Find:

The square root of the decimal number 12.25.

---

Solution:

We will use the long division method to find the square root of 12.25.

1. Place the decimal point in the quotient directly above the decimal point in the number 12.25.

2. Group the digits in pairs, starting from the decimal point. The integer part 12 forms the pair $\overline{12}$. The decimal part 25 forms the pair $\overline{25}$. So we have $\overline{12} . \overline{25}$.

3. Perform the long division:

$\begin{array}{c|cc} & 3 \ . \ 5 & \\ \hline 3 & \overline{12} \; . \overline{25} \\ + \; 3 & 9\phantom{(.)} \\ \hline 6 \; 5 & 3 \; 25 \\ + \; 5 & 3 \; 25 \\ \hline 70 & \phantom{(..)}0 \\ \end{array}$

Explanation of Steps:

a. Find the largest number whose square is less than or equal to the first pair (12). This is 3 ($3^2 = 9$). Write 3 as the divisor and the first digit of the quotient.

b. Subtract the square ($9$) from the first pair ($12 - 9 = 3$).

c. Bring down the next pair ($\overline{25}$) next to the remainder 3 to form the new dividend, 325.

d. Since we brought down the pair after the decimal point, place the decimal point in the quotient after 3.

e. Double the current quotient (3) to get $2 \times 3 = 6$. Find the largest digit $y$ such that the number formed by placing $y$ next to 6 (i.e., $6y$) multiplied by $y$ is less than or equal to the new dividend (325). If we try $y=5$, we get $65 \times 5 = 325$. This works exactly.

f. Write 5 as the next digit in the quotient (making it 3.5) and also place 5 next to 6 in the divisor place (making it 65).

g. Subtract the product ($65 \times 5 = 325$) from the dividend ($325 - 325 = 0$).

h. The remainder is 0, so the division is complete.

The square root obtained is 3.5.

---

Final Answer:

The square root of 12.25 is 3.5.

Given:

The area of the square plot = 2304 m².

---

To Find:

The length of the side of the square plot.

---

Solution:

Let the length of the side of the square plot be $s$ meters.

The formula for the area of a square is:

Area = $(\text{Side})^2 = s^2$

We are given that the area is 2304 m².

So, $s^2 = 2304$.

To find the side length $s$, we need to calculate the square root of 2304.

$s = \sqrt{2304}$

We can find the square root using the prime factorization method or the long division method.

Using Prime Factorization:

$\begin{array}{c|cc} 2 & 2304 \\ \hline 2 & 1152 \\ \hline 2 & 576 \\ \hline 2 & 288 \\ \hline 2 & 144 \\ \hline 2 & 72 \\ \hline 2 & 36 \\ \hline 2 & 18 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

The prime factorization is $2304 = 2 \times 2 \times 2 \times 2 \times 2 \times \ $$ 2 \times 2 \times 2 \ $$ \times 3 \times 3$.

Grouping the factors in pairs:

$2304 = (2 \times 2) \times (2 \times 2) \times (2 \times 2) \times (2 \times 2) \times (3 \times 3)$

Taking one factor from each pair:

$s = \sqrt{2304} = 2 \times 2 \times 2 \times 2 \times 3$

$s = 16 \times 3$

$s = 48$

Using Long Division Method:

$\begin{array}{c|cc} & 4 \ 8 & \\ \hline 4 & \overline{23} \; \overline{04} \\ + \; 4 & 16\phantom{()} \\ \hline 8 \; 8 & 7 \; 04 \\ + \; 8 & 7 \; 04 \\ \hline 96 & \phantom{00}0 \\ \end{array}$

The square root of 2304 is 48.

Since the side length must be positive, $s = 48$.

The side of the square plot is 48 meters.

---

Final Answer:

The side of the square plot is 48 m.

Given:

Total number of students = 2401.

Arrangement: Students stand in rows and columns.

Condition: The number of rows is equal to the number of columns.

---

To Find:

The number of rows.

---

Solution:

Let the number of rows be $x$.

Since the number of rows is equal to the number of columns, the number of columns is also $x$.

The total number of students is the product of the number of rows and the number of columns.

Total students = (Number of rows) $\times$ (Number of columns)

Total students = $x \times x$

Total students = $x^2$.

We are given that the total number of students is 2401.

Therefore, $x^2 = 2401$.

To find the number of rows, $x$, we need to find the square root of 2401.

$x = \sqrt{2401}$

We can find the square root using the prime factorization method:

$\begin{array}{c|cc} 7 & 2401 \\ \hline 7 & 343 \\ \hline 7 & 49 \\ \hline 7 & 7 \\ \hline & 1 \end{array}$

The prime factorization of 2401 is $7 \times 7 \times 7 \times 7 = 7^4$.

Grouping the factors into pairs:

$2401 = (7 \times 7) \times (7 \times 7)$

Taking one factor from each pair:

$x = \sqrt{2401} = 7 \times 7 = 49$.

Alternatively, using the long division method:

$\begin{array}{c|cc} & 4 \ 9 & \\ \hline 4 & \overline{24} \; \overline{01} \\ + \; 4 & 16\phantom{()} \\ \hline 8 \; 9 & 8 \; 01 \\ + \; 9 & 8 \; 01 \\ \hline 98 & 0 \end{array}$

The square root of 2401 is 49.

So, $x=49$.

The number of rows is 49.

---

Final Answer:

The number of rows is 49.

We will find the square root of each number using the Long Division Method.

---

(i) 2304

$\begin{array}{c|cc} & 4 \ 8 & \\ \hline 4 & \overline{23} \; \overline{04} \\ + \; 4 & 16\phantom{()} \\ \hline 8 \; 8 & 7 \; 04 \\ + \; 8 & 7 \; 04 \\ \hline 96 & \phantom{00}0 \\ \end{array}$

Therefore, $\sqrt{2304} = \mathbf{48}$.

---

(ii) 4489

$\begin{array}{c|cc} & 6 \ 7 & \\ \hline 6 & \overline{44} \; \overline{89} \\ + \; 6 & 36\phantom{()} \\ \hline 12 \; 7 & 8 \; 89 \\ + \; 7 & 8 \; 89 \\ \hline 134 & \phantom{00}0 \\ \end{array}$

Therefore, $\sqrt{4489} = \mathbf{67}$.

---

(iii) 3481

$\begin{array}{c|cc} & 5 \ 9 & \\ \hline 5 & \overline{34} \; \overline{81} \\ + \; 5 & 25\phantom{()} \\ \hline 10 \; 9 & 9 \; 81 \\ + \; 9 & 9 \; 81 \\ \hline 118 & \phantom{00}0 \\ \end{array}$

Therefore, $\sqrt{3481} = \mathbf{59}$.

---

(iv) 529

$\begin{array}{c|cc} & 2 \ 3 & \\ \hline 2 & \overline{5} \; \overline{29} \\ + \; 2 & 4\phantom{()} \\ \hline 4 \; 3 & 1 \; 29 \\ + \; 3 & 1 \; 29 \\ \hline 46 & \phantom{00}0 \\ \end{array}$

Therefore, $\sqrt{529} = \mathbf{23}$.

---

(v) 3249

$\begin{array}{c|cc} & 5 \ 7 & \\ \hline 5 & \overline{32} \; \overline{49} \\ + \; 5 & 25\phantom{()} \\ \hline 10 \; 7 & 7 \; 49 \\ + \; 7 & 7 \; 49 \\ \hline 114 & \phantom{00}0 \\ \end{array}$

Therefore, $\sqrt{3249} = \mathbf{57}$.

---

(vi) 1369

$\begin{array}{c|cc} & 3 \ 7 & \\ \hline 3 & \overline{13} \; \overline{69} \\ + \; 3 & 9\phantom{()} \\ \hline 6 \; 7 & 4 \; 69 \\ + \; 7 & 4 \; 69 \\ \hline 74 & \phantom{00}0 \\ \end{array}$

Therefore, $\sqrt{1369} = \mathbf{37}$.

---

(vii) 5776

$\begin{array}{c|cc} & 7 \ 6 & \\ \hline 7 & \overline{57} \; \overline{76} \\ + \; 7 & 49\phantom{()} \\ \hline 14 \; 6 & 8 \; 76 \\ + \; 6 & 8 \; 76 \\ \hline 152 & \phantom{00}0 \\ \end{array}$

Therefore, $\sqrt{5776} = \mathbf{76}$.

---

(viii) 7921

$\begin{array}{c|cc} & 8 \ 9 & \\ \hline 8 & \overline{79} \; \overline{21} \\ + \; 8 & 64\phantom{()} \\ \hline 16 \; 9 & 15 \; 21 \\ + \; 9 & 15 \; 21 \\ \hline 178 & \phantom{000}0 \\ \end{array}$

Therefore, $\sqrt{7921} = \mathbf{89}$.

---

(ix) 576

$\begin{array}{c|cc} & 2 \ 4 & \\ \hline 2 & \overline{5} \; \overline{76} \\ + \; 2 & 4\phantom{()} \\ \hline 4 \; 4 & 1 \; 76 \\ + \; 4 & 1 \; 76 \\ \hline 48 & \phantom{00}0 \\ \end{array}$

Therefore, $\sqrt{576} = \mathbf{24}$.

---

(x) 1024

$\begin{array}{c|cc} & 3 \ 2 & \\ \hline 3 & \overline{10} \; \overline{24} \\ + \; 3 & 9\phantom{()} \\ \hline 6 \; 2 & 1 \; 24 \\ + \; 2 & 1 \; 24 \\ \hline 64 & \phantom{00}0 \\ \end{array}$

Therefore, $\sqrt{1024} = \mathbf{32}$.

---

(xi) 3136

$\begin{array}{c|cc} & 5 \ 6 & \\ \hline 5 & \overline{31} \; \overline{36} \\ + \; 5 & 25\phantom{()} \\ \hline 10 \; 6 & 6 \; 36 \\ + \; 6 & 6 \; 36 \\ \hline 112 & \phantom{00}0 \\ \end{array}$

Therefore, $\sqrt{3136} = \mathbf{56}$.

---

(xii) 900

$\begin{array}{c|cc} & 3 \ 0 & \\ \hline 3 & \overline{9} \; \overline{00} \\ + \; 3 & 9\phantom{()} \\ \hline 6 \; 0 & 0 \; 00 \\ + \; 0 & \phantom{0}00 \\ \hline 60 & \phantom{00}0 \\ \end{array}$

Therefore, $\sqrt{900} = \mathbf{30}$.

We can find the number of digits in the square root of a perfect square by placing bars over pairs of digits starting from the unit's place (rightmost digit). The number of bars obtained is equal to the number of digits in the square root.

Alternatively, if a number has $n$ digits:

• If $n$ is even, the number of digits in its square root is $\frac{n}{2}$.
• If $n$ is odd, the number of digits in its square root is $\frac{n+1}{2}$.

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(i) 64

Place bars on 64: $\overline{64}$.

There is one bar.

Alternatively, the number of digits $n=2$ (even). Number of digits in square root = $\frac{n}{2} = \frac{2}{2} = 1$.

Therefore, the square root of 64 has 1 digit.

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(ii) 144

Place bars on 144: $\overline{1} \; \overline{44}$.

There are two bars.

Alternatively, the number of digits $n=3$ (odd). Number of digits in square root = $\frac{n+1}{2} = \frac{3+1}{2} = \frac{4}{2} = 2$.

Therefore, the square root of 144 has 2 digits.

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(iii) 4489

Place bars on 4489: $\overline{44} \; \overline{89}$.

There are two bars.

Alternatively, the number of digits $n=4$ (even). Number of digits in square root = $\frac{n}{2} = \frac{4}{2} = 2$.

Therefore, the square root of 4489 has 2 digits.

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(iv) 27225

Place bars on 27225: $\overline{2} \; \overline{72} \; \overline{25}$.

There are three bars.

Alternatively, the number of digits $n=5$ (odd). Number of digits in square root = $\frac{n+1}{2} = \frac{5+1}{2} = \frac{6}{2} = 3$.

Therefore, the square root of 27225 has 3 digits.

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(v) 390625

Place bars on 390625: $\overline{39} \; \overline{06} \; \overline{25}$.

There are three bars.

Alternatively, the number of digits $n=6$ (even). Number of digits in square root = $\frac{n}{2} = \frac{6}{2} = 3$.

Therefore, the square root of 390625 has 3 digits.

We will use the Long Division Method to find the square root of these decimal numbers.

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(i) 2.56

We pair the digits starting from the decimal point: $\overline{2} . \overline{56}$.

$\begin{array}{c|cc} & 1 \ . \ 6 & \\ \hline 1 & \overline{2} \; . \overline{56} \\ + \; 1 & 1\phantom{(.)} \\ \hline 2 \; 6 & 1 \; 56 \\ + \; 6 & 1 \; 56 \\ \hline 32 & \phantom{(..)}0 \\ \end{array}$

Therefore, $\sqrt{2.56} = \mathbf{1.6}$.

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(ii) 7.29

We pair the digits starting from the decimal point: $\overline{7} . \overline{29}$.

$\begin{array}{c|cc} & 2 \ . \ 7 & \\ \hline 2 & \overline{7} \; . \overline{29} \\ + \; 2 & 4\phantom{(.)} \\ \hline 4 \; 7 & 3 \; 29 \\ + \; 7 & 3 \; 29 \\ \hline 54 & \phantom{(..)}0 \\ \end{array}$

Therefore, $\sqrt{7.29} = \mathbf{2.7}$.

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(iii) 51.84

We pair the digits starting from the decimal point: $\overline{51} . \overline{84}$.

$\begin{array}{c|cc} & 7 \ . \ 2 & \\ \hline 7 & \overline{51} \; . \overline{84} \\ + \; 7 & 49\phantom{(.)} \\ \hline 14 \; 2 & 2 \; 84 \\ + \; 2 & 2 \; 84 \\ \hline 144 & \phantom{(..)}0 \\ \end{array}$

Therefore, $\sqrt{51.84} = \mathbf{7.2}$.

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(iv) 42.25

We pair the digits starting from the decimal point: $\overline{42} . \overline{25}$.

$\begin{array}{c|cc} & 6 \ . \ 5 & \\ \hline 6 & \overline{42} \; . \overline{25} \\ + \; 6 & 36\phantom{(.)} \\ \hline 12 \; 5 & 6 \; 25 \\ + \; 5 & 6 \; 25 \\ \hline 130 & \phantom{(..)}0 \\ \end{array}$

Therefore, $\sqrt{42.25} = \mathbf{6.5}$.

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(v) 31.36

We pair the digits starting from the decimal point: $\overline{31} . \overline{36}$.

$\begin{array}{c|cc} & 5 \ . \ 6 & \\ \hline 5 & \overline{31} \; . \overline{36} \\ + \; 5 & 25\phantom{(.)} \\ \hline 10 \; 6 & 6 \; 36 \\ + \; 6 & 6 \; 36 \\ \hline 112 & \phantom{(..)}0 \\ \end{array}$

Therefore, $\sqrt{31.36} = \mathbf{5.6}$.

To solve these problems, we use the long division method to find the square root of the given number. The remainder obtained during the division is the least number that must be subtracted from the original number to make it a perfect square. The quotient obtained is the square root of the resulting perfect square.

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(i) 402

Find the square root of 402 using long division:

$\begin{array}{c|cc} & 2 \ 0 & \\ \hline 2 & \overline{4} \; \overline{02} \\ + \; 2 & 4\phantom{()} \\ \hline 4 \; 0 & 0 \; 02 \\ + \; 0 & \phantom{0}00 \\ \hline 40 & \phantom{00}2 \\ \end{array}$

The quotient is 20 and the remainder is 2.

The least number to be subtracted is the remainder, which is 2.

The perfect square obtained is $402 - 2 = \mathbf{400}$.

The square root of the perfect square 400 is the quotient, which is $\sqrt{400} = \mathbf{20}$.

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(ii) 1989

Find the square root of 1989 using long division:

$\begin{array}{c|cc} & 4 \ 4 & \\ \hline 4 & \overline{19} \; \overline{89} \\ + \; 4 & 16\phantom{()} \\ \hline 8 \; 4 & 3 \; 89 \\ + \; 4 & 3 \; 36 \\ \hline 88 & \phantom{0}53 \\ \end{array}$

The quotient is 44 and the remainder is 53.

The least number to be subtracted is the remainder, which is 53.

The perfect square obtained is $1989 - 53 = \mathbf{1936}$.

The square root of the perfect square 1936 is the quotient, which is $\sqrt{1936} = \mathbf{44}$.

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(iii) 3250

Find the square root of 3250 using long division:

$\begin{array}{c|cc} & 5 \ 7 & \\ \hline 5 & \overline{32} \; \overline{50} \\ + \; 5 & 25\phantom{()} \\ \hline 10 \; 7 & 7 \; 50 \\ + \; 7 & 7 \; 49 \\ \hline 114 & \phantom{00}1 \\ \end{array}$

The quotient is 57 and the remainder is 1.

The least number to be subtracted is the remainder, which is 1.

The perfect square obtained is $3250 - 1 = \mathbf{3249}$.

The square root of the perfect square 3249 is the quotient, which is $\sqrt{3249} = \mathbf{57}$.

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(iv) 825

Find the square root of 825 using long division:

$\begin{array}{c|cc} & 2 \ 8 & \\ \hline 2 & \overline{8} \; \overline{25} \\ + \; 2 & 4\phantom{()} \\ \hline 4 \; 8 & 4 \; 25 \\ + \; 8 & 3 \; 84 \\ \hline 56 & \phantom{0}41 \\ \end{array}$

The quotient is 28 and the remainder is 41.

The least number to be subtracted is the remainder, which is 41.

The perfect square obtained is $825 - 41 = \mathbf{784}$.

The square root of the perfect square 784 is the quotient, which is $\sqrt{784} = \mathbf{28}$.

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(v) 4000

Find the square root of 4000 using long division:

$\begin{array}{c|cc} & 6 \ 3 & \\ \hline 6 & \overline{40} \; \overline{00} \\ + \; 6 & 36\phantom{()} \\ \hline 12 \; 3 & 4 \; 00 \\ + \; 3 & 3 \; 69 \\ \hline 126 & \phantom{0}31 \\ \end{array}$

The quotient is 63 and the remainder is 31.

The least number to be subtracted is the remainder, which is 31.

The perfect square obtained is $4000 - 31 = \mathbf{3969}$.

The square root of the perfect square 3969 is the quotient, which is $\sqrt{3969} = \mathbf{63}$.

To find the least number to be added, we first find the square root of the given number using the long division method. Let the quotient be $n$. The next perfect square is $(n+1)^2$. The number to be added is $(n+1)^2 - (\text{given number})$. The resulting perfect square is $(n+1)^2$, and its square root is $(n+1)$.

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(i) 525

Find the square root of 525 using long division:

$\begin{array}{c|cc} & 2 \ 2 & \\ \hline 2 & \overline{5} \; \overline{25} \\ + \; 2 & 4\phantom{()} \\ \hline 4 \; 2 & 1 \; 25 \\ + \; 2 & \phantom{0}84 \\ \hline 44 & \phantom{0}41 \\ \end{array}$

The quotient is 22. The given number 525 is greater than $22^2 = 484$.

The next perfect square is $(22+1)^2 = 23^2 = 529$.

The least number to be added = Next perfect square - Given number

Number to be added = $529 - 525 = \mathbf{4}$.

The perfect square obtained is $525 + 4 = \mathbf{529}$.

The square root of the perfect square 529 is $\sqrt{529} = \mathbf{23}$.

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(ii) 1750

Find the square root of 1750 using long division:

$\begin{array}{c|cc} & 4 \ 1 & \\ \hline 4 & \overline{17} \; \overline{50} \\ + \; 4 & 16\phantom{()} \\ \hline 8 \; 1 & 1 \; 50 \\ + \; 1 & \phantom{0}81 \\ \hline 82 & \phantom{0}69 \\ \end{array}$

The quotient is 41. The given number 1750 is greater than $41^2 = 1681$.

The next perfect square is $(41+1)^2 = 42^2 = 1764$.

The least number to be added = Next perfect square - Given number

Number to be added = $1764 - 1750 = \mathbf{14}$.

The perfect square obtained is $1750 + 14 = \mathbf{1764}$.

The square root of the perfect square 1764 is $\sqrt{1764} = \mathbf{42}$.

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(iii) 252

Find the square root of 252 using long division:

$\begin{array}{c|cc} & 1 \ 5 & \\ \hline 1 & \overline{2} \; \overline{52} \\ + \; 1 & 1\phantom{()} \\ \hline 2 \; 5 & 1 \; 52 \\ + \; 5 & 1 \; 25 \\ \hline 30 & \phantom{0}27 \\ \end{array}$

The quotient is 15. The given number 252 is greater than $15^2 = 225$.

The next perfect square is $(15+1)^2 = 16^2 = 256$.

The least number to be added = Next perfect square - Given number

Number to be added = $256 - 252 = \mathbf{4}$.

The perfect square obtained is $252 + 4 = \mathbf{256}$.

The square root of the perfect square 256 is $\sqrt{256} = \mathbf{16}$.

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(iv) 1825

Find the square root of 1825 using long division:

$\begin{array}{c|cc} & 4 \ 2 & \\ \hline 4 & \overline{18} \; \overline{25} \\ + \; 4 & 16\phantom{()} \\ \hline 8 \; 2 & 2 \; 25 \\ + \; 2 & 1 \; 64 \\ \hline 84 & \phantom{0}61 \\ \end{array}$

The quotient is 42. The given number 1825 is greater than $42^2 = 1764$.

The next perfect square is $(42+1)^2 = 43^2 = 1849$.

The least number to be added = Next perfect square - Given number

Number to be added = $1849 - 1825 = \mathbf{24}$.

The perfect square obtained is $1825 + 24 = \mathbf{1849}$.

The square root of the perfect square 1849 is $\sqrt{1849} = \mathbf{43}$.

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(v) 6412

Find the square root of 6412 using long division:

$\begin{array}{c|cc} & 8 \ 0 & \\ \hline 8 & \overline{64} \; \overline{12} \\ + \; 8 & 64\phantom{()} \\ \hline 16 \; 0 & 0 \; 12 \\ + \; 0 & \phantom{00}0 \\ \hline 160 & \phantom{0}12 \\ \end{array}$

The quotient is 80. The given number 6412 is greater than $80^2 = 6400$.

The next perfect square is $(80+1)^2 = 81^2 = 6561$.

The least number to be added = Next perfect square - Given number

Number to be added = $6561 - 6412 = \mathbf{149}$.

The perfect square obtained is $6412 + 149 = \mathbf{6561}$.

The square root of the perfect square 6561 is $\sqrt{6561} = \mathbf{81}$.

Given:

The area of the square = 441 m².

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To Find:

The length of the side of the square.

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Solution:

Let the length of the side of the square be $s$ meters.

The formula for the area of a square is given by:

Area = $(\text{Side})^2$

Area = $s^2$

We are given that the area is 441 m².

So, $s^2 = 441$.

To find the length of the side, $s$, we need to calculate the square root of 441.

$s = \sqrt{441}$

We can find the square root using the prime factorization method or the long division method.

Using Prime Factorization:

$\begin{array}{c|cc} 3 & 441 \\ \hline 3 & 147 \\ \hline 7 & 49 \\ \hline 7 & 7 \\ \hline & 1 \end{array}$

$441 = 3 \times 3 \times 7 \times 7 = (3 \times 3) \times (7 \times 7)$

Taking one factor from each pair:

$s = \sqrt{441} = 3 \times 7 = 21$.

Using Long Division Method:

$\begin{array}{c|cc} & 2 \ 1 & \\ \hline 2 & \overline{4} \; \overline{41} \\ + \; 2 & 4\phantom{()} \\ \hline 4 \; 1 & 0 \; 41 \\ + \; 1 & \phantom{0}41 \\ \hline 42 & \phantom{00}0 \\ \end{array}$

The square root of 441 is 21.

Since the side length must be positive, $s = 21$.

The length of the side of the square is 21 meters.

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Final Answer:

The length of the side of the square is 21 m.

In a right-angled triangle ABC, where $\angle B = 90^\circ$, AC is the hypotenuse, and AB and BC are the legs. We can use the Pythagorean theorem, which states:

$AC^2 = AB^2 + BC^2$

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(a) If AB = 6 cm, BC = 8 cm, find AC

Given:

In $\triangle ABC$, $\angle B = 90^\circ$.

AB = 6 cm

BC = 8 cm

To Find:

The length of AC.

Solution:

Using the Pythagorean theorem:

$AC^2 = AB^2 + BC^2$

Substitute the given values:

$AC^2 = 6^2 + 8^2$

$AC^2 = 36 + 64$

$AC^2 = 100$

To find AC, take the square root of both sides:

$AC = \sqrt{100}$

$AC = 10$

Since length must be positive, AC = 10 cm.

Result (a): The length of AC is 10 cm.

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(b) If AC = 13 cm, BC = 5 cm, find AB

Given:

In $\triangle ABC$, $\angle B = 90^\circ$.

AC = 13 cm (Hypotenuse)

BC = 5 cm

To Find:

The length of AB.

Solution:

Using the Pythagorean theorem:

$AC^2 = AB^2 + BC^2$

Rearrange the formula to solve for AB:

$AB^2 = AC^2 - BC^2$

Substitute the given values:

$AB^2 = 13^2 - 5^2$

$AB^2 = 169 - 25$

$AB^2 = 144$

To find AB, take the square root of both sides:

$AB = \sqrt{144}$

$AB = 12$

Since length must be positive, AB = 12 cm.

Result (b): The length of AB is 12 cm.

Given:

Total number of plants the gardener has = 1000.

Desired arrangement: Number of rows = Number of columns.

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To Find:

The minimum number of additional plants required to achieve the desired arrangement.

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Solution:

Let the number of rows be $x$.

Since the number of rows must equal the number of columns, the number of columns is also $x$.

The total number of plants needed for this square arrangement is $x \times x = x^2$.

This means the total number of plants must be a perfect square.

The gardener has 1000 plants. We need to find the smallest perfect square number that is greater than 1000.

First, let's find the approximate square root of 1000 using the long division method.

$\begin{array}{c|cc} & 3 \ 1 & \\ \hline 3 & \overline{10} \; \overline{00} \\ + \; 3 & 9\phantom{()} \\ \hline 6 \; 1 & 1 \; 00 \\ + \; 1 & \phantom{0}61 \\ \hline 62 & \phantom{0}39 \\ \end{array}$

The long division shows that the largest integer whose square is less than 1000 is 31 ($31^2 = 961$).

Since 1000 is not a perfect square (remainder is 39), the gardener cannot arrange exactly 1000 plants in the desired way.

The next integer after 31 is $31 + 1 = 32$.

The next perfect square is $32^2$.

$32^2 = 32 \times 32 = 1024$.

So, the minimum number of plants required for the square arrangement is 1024.

The number of additional plants needed is the difference between the required perfect square number and the number of plants the gardener currently has.

Minimum number of plants needed more = (Required perfect square) - (Plants available)

Minimum number of plants needed more = $1024 - 1000$

Minimum number of plants needed more = 24.

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Final Answer:

The minimum number of plants the gardener needs more is 24.

Given:

Total number of children = 500.

Required arrangement: To stand in rows and columns such that the number of rows equals the number of columns.

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To Find:

The number of children who would be left out of this arrangement.

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Solution:

Let the number of rows be $x$.

Since the number of rows must equal the number of columns, the number of columns is also $x$.

The total number of children that can be arranged in this manner is $x \times x = x^2$.

This means the number of children arranged must be a perfect square.

We need to find the largest perfect square number less than or equal to 500. The difference between 500 and this perfect square will be the number of children left out.

We use the long division method to find the square root of 500 and the remainder.

$\begin{array}{c|cc} & 2 \ 2 & \\ \hline 2 & \overline{5} \; \overline{00} \\ + \; 2 & 4\phantom{()} \\ \hline 4 \; 2 & 1 \; 00 \\ + \; 2 & \phantom{0}84 \\ \hline 44 & \phantom{0}16 \\ \end{array}$

The quotient is 22 and the remainder is 16.

This means that the largest perfect square less than 500 is $22^2 = 484$.

So, 484 children can be arranged in 22 rows and 22 columns.

The number of children left out is the remainder obtained from the division.

Number of children left out = Total children - Number of children arranged

Number of children left out = $500 - 22^2$

Number of children left out = $500 - 484$

Number of children left out = 16.

Alternatively, the remainder from the long division directly gives the number of children left out, which is 16.

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Final Answer:

16 children would be left out in this arrangement.