Squares and Square Roots, Cubes and Cube Roots — Exercise 3(A)

Find the square of the following number:

14

Answer

The square of a number is that number raised to the power 2.

⇒ (14)²

⇒ 14 × 14

⇒ 196

Hence, (14)² = 196.

Find the square of the following number:

137

Answer

The square of a number is that number raised to the power 2.

⇒ (137)²

⇒ 137 × 137

⇒ 18769

Hence, (137)² = 18769.

Find the square of the following number:

$\dfrac{4}{17}$

Answer

The square of a number is that number raised to the power 2.

$\Rightarrow \left(\dfrac{4}{17}\right)^2\\[1em] \Rightarrow \dfrac{4}{17} \times \dfrac{4}{17}\\[1em] \Rightarrow \dfrac{16}{289}$

Hence, $\mathbf{\left(\dfrac{4}{17}\right)^2 = \dfrac{16}{289}}$.

Find the square of the following number:

$2\dfrac{3}{4}$

Answer

Converting mixed fraction into improper fraction:

$2\dfrac{3}{4} = \dfrac{11}{4}$

The square of a number is that number raised to the power 2.

$\Rightarrow \left(2\dfrac{3}{4}\right)^2\\[1em] \Rightarrow \left(\dfrac{11}{4}\right)^2\\[1em] \Rightarrow \dfrac{11}{4} \times \dfrac{11}{4}\\[1em] \Rightarrow \dfrac{121}{16}\\[1em] \Rightarrow 7\dfrac{9}{16}$

Hence, $\mathbf{\left(2\dfrac{3}{4}\right)^2 = 7\dfrac{9}{16}}$.

Find the square of the following number:

0.01

Answer

The square of a number is that number raised to the power 2.

⇒ (0.01)²

⇒ 0.01 × 0.01

⇒ 0.0001

Hence, (0.01)² = 0.0001.

Find the square of the following number:

1.2

Answer

The square of a number is that number raised to the power 2.

⇒ (1.2)²

⇒ 1.2 × 1.2

⇒ 1.44

Hence, (1.2)² = 1.44.

Find the square of the following number:

0.17

Answer

The square of a number is that number raised to the power 2.

⇒ (0.17)²

⇒ 0.17 × 0.17

⇒ 0.0289

Hence, (0.17)² = 0.0289.

Find the square of the following number:

4.6

Answer

The square of a number is that number raised to the power 2.

⇒ (4.6)²

⇒ 4.6 × 4.6

⇒ 21.16

Hence, (4.6)² = 21.16.

Using prime factorization method, find whether the following number is a perfect square number:

252

Answer

Resolving 252 into prime factors, we get :

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

252 = (2 × 2) × (3 × 3) × 7

Since the prime factor 7 is left unpaired, so 252 cannot be expressed as the product of pairs of equal prime factors.

Hence, 252 is not a perfect square number.

Using prime factorization method, find whether the following number is a perfect square number:

324

Answer

Resolving 324 into prime factors, we get :

$\begin{array}{l|r} 2 & 324 \\ \hline 2 & 162 \\ \hline 3 & 81 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \\ \end{array}$

324 = (2 × 2) × (3 × 3) × (3 × 3)

Since 324 can be expressed as the product of pairs of equal prime factors.

Hence, 324 is a perfect square number.

Using prime factorization method, find whether the following number is a perfect square number:

676

Answer

Resolving 676 into prime factors, we get :

$\begin{array}{l|r} 2 & 676 \\ \hline 2 & 338 \\ \hline 13 & 169 \\ \hline 13 & 13 \\ \hline & 1 \\ \end{array}$

676 = (2 × 2) × (13 × 13)

Since 676 can be expressed as the product of pairs of equal prime factors.

Hence, 676 is a perfect square number.

Using prime factorization method, find whether the following number is a perfect square number:

1225

Answer

Resolving 1225 into prime factors, we get :

$\begin{array}{l|r} 5 & 1225 \\ \hline 5 & 245 \\ \hline 7 & 49 \\ \hline 7 & 7 \\ \hline & 1 \\ \end{array}$

1225 = (5 × 5) × (7 × 7)

Since 1225 can be expressed as the product of pairs of equal prime factors.

Hence, 1225 is a perfect square number.

Using prime factorization method, find whether the following number is a perfect square number:

2916

Answer

Resolving 2916 into prime factors, we get :

$\begin{array}{l|r} 2 & 2916 \\ \hline 2 & 1458 \\ \hline 3 & 729 \\ \hline 3 & 243 \\ \hline 3 & 81 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \\ \end{array}$

2916 = (2 × 2) × (3 × 3) × (3 × 3) × (3 × 3)

Since 2916 can be expressed as the product of pairs of equal prime factors.

Hence, 2916 is a perfect square number.

Using prime factorization method, find whether the following number is a perfect square number:

3528

Answer

Resolving 3528 into prime factors, we get :

$\begin{array}{l|r} 2 & 3528 \\ \hline 2 & 1764 \\ \hline 2 & 882 \\ \hline 3 & 441 \\ \hline 3 & 147 \\ \hline 7 & 49 \\ \hline 7 & 7 \\ \hline & 1 \\ \end{array}$

3528 = (2 × 2) × (3 × 3) × (7 × 7) × 2

Since the prime factor 2 is left unpaired, so 3528 cannot be expressed as the product of pairs of equal prime factors.

Hence, 3528 is not a perfect square number.

Using prime factorization method, find whether the following number is a perfect square number:

5625

Answer

Resolving 5625 into prime factors, we get :

$\begin{array}{l|r} 3 & 5625 \\ \hline 3 & 1875 \\ \hline 5 & 625 \\ \hline 5 & 125 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \\ \end{array}$

5625 = (3 × 3) × (5 × 5) × (5 × 5)

Since 5625 can be expressed as the product of pairs of equal prime factors.

Hence, 5625 is a perfect square number.

Using prime factorization method, find whether the following number is a perfect square number:

9075

Answer

Resolving 9075 into prime factors, we get :

$\begin{array}{l|r} 3 & 9075 \\ \hline 5 & 3025 \\ \hline 5 & 605 \\ \hline 11 & 121 \\ \hline 11 & 11 \\ \hline & 1 \\ \end{array}$

9075 = (5 × 5) × (11 × 11) × 3

Since the prime factor 3 is left unpaired, so 9075 cannot be expressed as the product of pairs of equal prime factors.

Hence, 9075 is not a perfect square number.

Show that 441 is a perfect square. Also, find the square root of it.

Answer

Resolving 441 into prime factors, we get :

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

441 = (3 × 3) × (7 × 7)

Since 441 can be expressed as the product of pairs of equal prime factors, so 441 is a perfect square.

$\Rightarrow \sqrt{441} = \sqrt{(3 × 3) × (7 × 7)} \\[1em] \Rightarrow \sqrt{441} = 3 \times 7\\[1em] \Rightarrow \sqrt{441} = 21$

Hence, 441 is a perfect square and $\mathbf{\sqrt{441} = 21}$.

Show that 784 is a perfect square. Also, find the square root of it.

Answer

Resolving 784 into prime factors, we get :

$\begin{array}{l|r} 2 & 784 \\ \hline 2 & 392 \\ \hline 2 & 196 \\ \hline 2 & 98 \\ \hline 7 & 49 \\ \hline 7 & 7 \\ \hline & 1 \\ \end{array}$

784 = (2 × 2) × (2 × 2) × (7 × 7)

Since 784 can be expressed as the product of pairs of equal prime factors, so 784 is a perfect square.

$\Rightarrow \sqrt{784} = \sqrt{(2 \times 2) \times (2 \times 2) \times (7 \times 7)} \\[1em] \Rightarrow \sqrt{784} = 2 \times 2 \times 7\\[1em] \Rightarrow \sqrt{784} = 28$

Hence, 784 is a perfect square and $\mathbf{\sqrt{784} = 28}$.

Show that 1225 is a perfect square. Also, find the square root of it.

Answer

Resolving 1225 into prime factors, we get :

$\begin{array}{l|r} 5 & 1225 \\ \hline 5 & 245 \\ \hline 7 & 49 \\ \hline 7 & 7 \\ \hline & 1 \\ \end{array}$

1225 = (5 × 5) × (7 × 7)

Since 1225 can be expressed as the product of pairs of equal prime factors, so 1225 is a perfect square.

$\Rightarrow \sqrt{1225} = \sqrt{(5 \times 5) \times (7 \times 7)} \\[1em] \Rightarrow \sqrt{1225} = 5 \times 7\\[1em] \Rightarrow \sqrt{1225} = 35$

Hence, 1225 is a perfect square and $\mathbf{\sqrt{1225} = 35}$.

Show that 3969 is a perfect square. Also, find the square root of it.

Answer

Resolving 3969 into prime factors, we get :

$\begin{array}{l|r} 3 & 3969 \\ \hline 3 & 1323 \\ \hline 3 & 441 \\ \hline 3 & 147 \\ \hline 7 & 49 \\ \hline 7 & 7 \\ \hline & 1 \\ \end{array}$

3969 = (3 × 3) × (3 × 3) × (7 × 7)

Since 3969 can be expressed as the product of pairs of equal prime factors, so 3969 is a perfect square.

$\Rightarrow \sqrt{3969} = \sqrt{(3 \times 3) \times (3 \times 3) \times (7 \times 7)} \\[1em] \Rightarrow \sqrt{3969} = 3 \times 3 \times 7\\[1em] \Rightarrow \sqrt{3969} = 63$

Hence, 3969 is a perfect square and $\mathbf{\sqrt{3969} = 63}$.

Show that 2601 is a perfect square. Also, find the square root of it.

Answer

Resolving 2601 into prime factors, we get :

$\begin{array}{l|r} 3 & 2601 \\ \hline 3 & 867 \\ \hline 17 & 289 \\ \hline 17 & 17 \\ \hline & 1 \\ \end{array}$

2601 = (3 × 3) × (17 × 17)

Since 2601 can be expressed as the product of pairs of equal prime factors, so 2601 is a perfect square.

$\Rightarrow \sqrt{2601} = \sqrt{(3 \times 3) \times (17 \times 17)} \\[1em] \Rightarrow \sqrt{2601} = 3 \times 17\\[1em] \Rightarrow \sqrt{2601} = 51$

Hence, 2601 is a perfect square and $\mathbf{\sqrt{2601} = 51}$.

Show that 5929 is a perfect square. Also, find the square root of it.

Answer

Resolving 5929 into prime factors, we get :

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

5929 = (7 × 7) × (11 × 11)

Since 5929 can be expressed as the product of pairs of equal prime factors, so 5929 is a perfect square.

$\Rightarrow \sqrt{5929} = \sqrt{(7 \times 7) \times (11 \times 11)} \\[1em] \Rightarrow \sqrt{5929} = 7 \times 11\\[1em] \Rightarrow \sqrt{5929} = 77$

Hence, 5929 is a perfect square and $\mathbf{\sqrt{5929} = 77}$.

Show that 7056 is a perfect square. Also, find the square root of it.

Answer

Resolving 7056 into prime factors, we get :

$\begin{array}{l|r} 2 & 7056 \\ \hline 2 & 3528 \\ \hline 2 & 1764 \\ \hline 2 & 882 \\ \hline 3 & 441 \\ \hline 3 & 147 \\ \hline 7 & 49 \\ \hline 7 & 7 \\ \hline & 1 \\ \end{array}$

7056 = (2 × 2) × (2 × 2) × (3 × 3) × (7 × 7)

Since 7056 can be expressed as the product of pairs of equal prime factors, so 7056 is a perfect square.

$\Rightarrow \sqrt{7056} = \sqrt{(2 \times 2) \times (2 \times 2) \times (3 \times 3) \times (7 \times 7)} \\[1em] \Rightarrow \sqrt{7056} = 2 \times 2 \times 3 \times 7\\[1em] \Rightarrow \sqrt{7056} = 84$

Hence, 7056 is a perfect square and $\mathbf{\sqrt{7056} = 84}$.

Show that 8281 is a perfect square. Also, find the square root of it.

Answer

Resolving 8281 into prime factors, we get :

$\begin{array}{l|r} 7 & 8281 \\ \hline 7 & 1183 \\ \hline 13 & 169 \\ \hline 13 & 13 \\ \hline & 1 \\ \end{array}$

8281 = (7 × 7) × (13 × 13)

Since 8281 can be expressed as the product of pairs of equal prime factors, so 8281 is a perfect square.

$\Rightarrow \sqrt{8281} = \sqrt{(7 \times 7) \times (13 \times 13)} \\[1em] \Rightarrow \sqrt{8281} = 7 \times 13\\[1em] \Rightarrow \sqrt{8281} = 91$

Hence, 8281 is a perfect square and $\mathbf{\sqrt{8281} = 91}$.

Find the smallest number by which 588 must be multiplied to get a perfect square number. Also, find the square root of the resulting number.

Answer

Resolving 588 into prime factors, we get :

$\begin{array}{l|r} 2 & 588 \\ \hline 2 & 294 \\ \hline 3 & 147 \\ \hline 7 & 49 \\ \hline 7 & 7 \\ \hline & 1 \\ \end{array}$

588 = (2 × 2) × (7 × 7) × 3

Thus, 2, 2 and 7, 7 exist in pairs while 3 is alone.

So, we should multiply the given number by 3 to get a perfect square number.

$\Rightarrow 588 \times 3 = 1764\\[1em] \Rightarrow 1764 = (2 \times 2) \times (3 \times 3) \times (7 \times 7)\\[1em] \Rightarrow \sqrt{1764} = \sqrt{(2 \times 2) \times (3 \times 3) \times (7 \times 7)} \\[1em] \Rightarrow \sqrt{1764} = 2 \times 3 \times 7\\[1em] \Rightarrow \sqrt{1764} = 42$

Hence, 588 should be multiplied by 3 and $\mathbf{\sqrt{1764} = 42}$.

Find the smallest number by which 2592 must be multiplied to get a perfect square number. Also, find the square root of the resulting number.

Answer

Resolving 2592 into prime factors, we get :

$\begin{array}{l|r} 2 & 2592 \\ \hline 2 & 1296 \\ \hline 2 & 648 \\ \hline 2 & 324 \\ \hline 2 & 162 \\ \hline 3 & 81 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \\ \end{array}$

2592 = (2 × 2) × (2 × 2) × (3 × 3) × (3 × 3) × 2

Thus, 2, 2 and 3, 3 exist in pairs while one 2 is alone.

So, we should multiply the given number by 2 to get a perfect square number.

$\Rightarrow 2592 \times 2 = 5184\\[1em] \Rightarrow 5184 = (2 \times 2) \times (2 \times 2) \times (2 \times 2) \times (3 \times 3) \times (3 \times 3)\\[1em] \Rightarrow \sqrt{5184} = \sqrt{(2 \times 2) \times (2 \times 2) \times (2 \times 2) \times (3 \times 3) \times (3 \times 3)} \\[1em] \Rightarrow \sqrt{5184} = 2 \times 2 \times 2 \times 3 \times 3\\[1em] \Rightarrow \sqrt{5184} = 72$

Hence, 2592 should be multiplied by 2 and $\mathbf{\sqrt{5184} = 72}$.

Find the smallest number by which 3332 must be multiplied to get a perfect square number. Also, find the square root of the resulting number.

Answer

Resolving 3332 into prime factors, we get :

$\begin{array}{l|r} 2 & 3332 \\ \hline 2 & 1666 \\ \hline 7 & 833 \\ \hline 7 & 119 \\ \hline 17 & 17 \\ \hline & 1 \\ \end{array}$

3332 = (2 × 2) × (7 × 7) × 17

Thus, 2, 2 and 7, 7 exist in pairs while 17 is alone.

So, we should multiply the given number by 17 to get a perfect square number.

$\Rightarrow 3332 \times 17 = 56644\\[1em] \Rightarrow 56644 = (2 \times 2) \times (7 \times 7) \times (17 \times 17)\\[1em] \Rightarrow \sqrt{56644} = \sqrt{(2 \times 2) \times (7 \times 7) \times (17 \times 17)} \\[1em] \Rightarrow \sqrt{56644} = 2 \times 7 \times 17\\[1em] \Rightarrow \sqrt{56644} = 238$

Hence, 3332 should be multiplied by 17 and $\mathbf{\sqrt{56644} = 238}$.

Find the smallest number by which 3380 must be multiplied to get a perfect square number. Also, find the square root of the resulting number.

Answer

Resolving 3380 into prime factors, we get :

$\begin{array}{l|r} 2 & 3380 \\ \hline 2 & 1690 \\ \hline 5 & 845 \\ \hline 13 & 169 \\ \hline 13 & 13 \\ \hline & 1 \\ \end{array}$

3380 = (2 × 2) × (13 × 13) × 5

Thus, 2, 2 and 13, 13 exist in pairs while 5 is alone.

So, we should multiply the given number by 5 to get a perfect square number.

$\Rightarrow 3380 \times 5 = 16900\\[1em] \Rightarrow 16900 = (2 \times 2) \times (5 \times 5) \times (13 \times 13)\\[1em] \Rightarrow \sqrt{16900} = \sqrt{(2 \times 2) \times (5 \times 5) \times (13 \times 13)} \\[1em] \Rightarrow \sqrt{16900} = 2 \times 5 \times 13\\[1em] \Rightarrow \sqrt{16900} = 130$

Hence, 3380 should be multiplied by 5 and $\mathbf{\sqrt{16900} = 130}$.

By which least number should 1728 be divided to obtain a perfect square number? Also, find the resulting perfect square number and its square root.

Answer

Resolving 1728 into prime factors, we get :

$\begin{array}{l|r} 2 & 1728 \\ \hline 2 & 864 \\ \hline 2 & 432 \\ \hline 2 & 216 \\ \hline 2 & 108 \\ \hline 2 & 54 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \\ \end{array}$

1728 = (2 × 2) × (2 × 2) × (2 × 2) × (3 × 3) × 3

Clearly, in the product of prime factors, the factor 3 does not exist in pair.

So, we must divide the given number by 3.

$\Rightarrow 1728 \div 3 = 576\\[1em] \Rightarrow 576 = (2 \times 2) \times (2 \times 2) \times (2 \times 2) \times (3 \times 3)\\[1em] \Rightarrow \sqrt{576} = \sqrt{(2 \times 2) \times (2 \times 2) \times (2 \times 2) \times (3 \times 3)} \\[1em] \Rightarrow \sqrt{576} = 2 \times 2 \times 2 \times 3\\[1em] \Rightarrow \sqrt{576} = 24$

Hence, 1728 should be divided by 3, the resulting perfect square number is 576 and $\mathbf{\sqrt{576} = 24}$.

By which least number should 4500 be divided to obtain a perfect square number? Also, find the resulting perfect square number and its square root.

Answer

Resolving 4500 into prime factors, we get :

$\begin{array}{l|r} 2 & 4500 \\ \hline 2 & 2250 \\ \hline 3 & 1125 \\ \hline 3 & 375 \\ \hline 5 & 125 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \\ \end{array}$

4500 = (2 × 2) × (3 × 3) × (5 × 5) × 5

Clearly, in the product of prime factors, the factor 5 does not exist in pair.

So, we must divide the given number by 5.

$\Rightarrow 4500 \div 5 = 900\\[1em] \Rightarrow 900 = (2 \times 2) \times (3 \times 3) \times (5 \times 5)\\[1em] \Rightarrow \sqrt{900} = \sqrt{(2 \times 2) \times (3 \times 3) \times (5 \times 5)} \\[1em] \Rightarrow \sqrt{900} = 2 \times 3 \times 5\\[1em] \Rightarrow \sqrt{900} = 30$

Hence, 4500 should be divided by 5, the resulting perfect square number is 900 and $\mathbf{\sqrt{900} = 30}$.

By which least number should 7776 be divided to obtain a perfect square number? Also, find the resulting perfect square number and its square root.

Answer

Resolving 7776 into prime factors, we get :

$\begin{array}{l|r} 2 & 7776 \\ \hline 2 & 3888 \\ \hline 2 & 1944 \\ \hline 2 & 972 \\ \hline 2 & 486 \\ \hline 3 & 243 \\ \hline 3 & 81 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \\ \end{array}$

7776 = (2 × 2) × (2 × 2) × (3 × 3) × (3 × 3) × 2 × 3

Clearly, in the product of prime factors, the factors 2 and 3 do not exist in pairs.

So, we must divide the given number by 2 × 3, i.e., 6.

$\Rightarrow 7776 \div 6 = 1296\\[1em] \Rightarrow 1296 = (2 \times 2) \times (2 \times 2) \times (3 \times 3) \times (3 \times 3)\\[1em] \Rightarrow \sqrt{1296} = \sqrt{(2 \times 2) \times (2 \times 2) \times (3 \times 3) \times (3 \times 3)} \\[1em] \Rightarrow \sqrt{1296} = 2 \times 2 \times 3 \times 3\\[1em] \Rightarrow \sqrt{1296} = 36$

Hence, 7776 should be divided by 6, the resulting perfect square number is 1296 and $\mathbf{\sqrt{1296} = 36}$.

By which least number should 8820 be divided to obtain a perfect square number? Also, find the resulting perfect square number and its square root.

Answer

Resolving 8820 into prime factors, we get :

$\begin{array}{l|r} 2 & 8820 \\ \hline 2 & 4410 \\ \hline 3 & 2205 \\ \hline 3 & 735 \\ \hline 5 & 245 \\ \hline 7 & 49 \\ \hline 7 & 7 \\ \hline & 1 \\ \end{array}$

8820 = (2 × 2) × (3 × 3) × (7 × 7) × 5

Clearly, in the product of prime factors, the factor 5 does not exist in pair.

So, we must divide the given number by 5.

$\Rightarrow 8820 \div 5 = 1764\\[1em] \Rightarrow 1764 = (2 \times 2) \times (3 \times 3) \times (7 \times 7)\\[1em] \Rightarrow \sqrt{1764} = \sqrt{(2 \times 2) \times (3 \times 3) \times (7 \times 7)} \\[1em] \Rightarrow \sqrt{1764} = 2 \times 3 \times 7\\[1em] \Rightarrow \sqrt{1764} = 42$

Hence, 8820 should be divided by 5, the resulting perfect square number is 1764 and $\mathbf{\sqrt{1764} = 42}$.

The students of a class arranged a picnic. Each student contributed as many rupees as the number of students in the class. If the total contribution is ₹ 1156, find the strength of the class.

Answer

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

Then, the amount contributed by each student = number of students = ₹$x$.

∴ Total contribution = $x \times x = x^2$.

But, total contribution = ₹1,156 (given).

Resolving 1156 into prime factors, we get :

$\begin{array}{l|r} 2 & 1156 \\ \hline 2 & 578 \\ \hline 17 & 289 \\ \hline 17 & 17 \\ \hline & 1 \\ \end{array}$

1156 = (2 × 2) × (17 × 17)

$\Rightarrow x^2 = 1156\\[1em] \Rightarrow x = \sqrt{1156}\\[1em] \Rightarrow x = \sqrt{(2 \times 2) \times (17 \times 17)}\\[1em] \Rightarrow x = 2 \times 17\\[1em] \Rightarrow x = 34$

Hence, the strength of the class is 34.

In an auditorium, the number of rows is equal to the number of chairs in each row. If the capacity of the auditorium is 2025, find the number of chairs in each row.

Answer

Let the number of rows be $x$.

Then, the number of chairs in each row = number of rows = $x$.

∴ Total number of chairs = $x \times x = x^2$.

But, total number of chairs = 2025 (given).

Resolving 2025 into prime factors, we get :

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

2025 = (3 × 3) × (3 × 3) × (5 × 5)

$\Rightarrow x^2 = 2025\\[1em] \Rightarrow x = \sqrt{2025}\\[1em] \Rightarrow x = \sqrt{(3 \times 3) \times (3 \times 3) \times (5 \times 5)}\\[1em] \Rightarrow x = 3 \times 3 \times 5\\[1em] \Rightarrow x = 45$

Hence, there are 45 chairs in each row.

Find the least square number which is exactly divisible by each of the numbers 8, 9, 10 and 15.

Answer

The least number divisible by each one of 8, 9, 10 and 15 is their L.C.M.

Finding the L.C.M. of 8, 9, 10 and 15 :

$\begin{array}{l|rrrr} 2 & 8 & 9 & 10 & 15 \\ \hline 2 & 4 & 9 & 5 & 15 \\ \hline 2 & 2 & 9 & 5 & 15 \\ \hline 3 & 1 & 9 & 5 & 15 \\ \hline 3 & 1 & 3 & 5 & 5 \\ \hline 5 & 1 & 1 & 5 & 5 \\ \hline & 1 & 1 & 1 & 1 \\ \end{array}$

L.C.M. of 8, 9, 10 and 15 = 2 × 2 × 2 × 3 × 3 × 5 = 360

By prime factorisation we get :

360 = (2 × 2) × (3 × 3) × 2 × 5

To make it a perfect square, it must be multiplied by 2 × 5, i.e., by 10.

$\Rightarrow 360 \times 10 = 3600\\[1em] \Rightarrow 3600 = (2 \times 2) \times (2 \times 2) \times (3 \times 3) \times (5 \times 5)\\[1em] \Rightarrow \sqrt{3600} = \sqrt{(2 \times 2) \times (2 \times 2) \times (3 \times 3) \times (5 \times 5)} \\[1em] \Rightarrow \sqrt{3600} = 2 \times 2 \times 3 \times 5\\[1em] \Rightarrow \sqrt{3600} = 60$

Hence, the least square number which is exactly divisible by each of the numbers 8, 9, 10 and 15 is 3600.