 # Show that each of the following numbers is a perfect square. `
Question:

Show that each of the following numbers is a perfect square. Also, find the number whose square is the given number in each case:

(i) 1156

(ii) 2025

(iii) 14641

(iv) 4761

Solution:

In each problem, factorise the number into its prime factors.

(i) 1156 = 2 x 2 x 17 x 17

Grouping the factors into pairs of equal factors, we obtain:

1156 = (2 x 2) x (17 x 17)

No factors are left over. Hence, 1156 is a perfect square. Moreover, by grouping 1156 into equal factors:

1156 = (2 x 17) x (2 x 17)

= (2 x 17)2
Hence, 1156 is the square of 34, which is equal to 2 x 17.

(ii) 2025 = 3 x 3 x 3 x 3 x 5 x 5

Grouping the factors into pairs of equal factors, we obtain:

2025 = (3 x 3) x (3 x 3) x (5 x 5)

No factors are left over. Hence, 2025 is a perfect square. Moreover, by grouping 2025 into equal factors:

2025 = (3 x 3 x 5) x (3 x 3 x 5)

= (3 x 3 x 5)2

Hence, 2025 is the square of 45, which is equal to 3 x 3 x 5.

(iii) 14641 = 11 x 11 x 11 x 11

Grouping the factors into pairs of equal factors, we obtain:

14641 = (11 x 11) x (11 x 11)

No factors are left over. Hence, 14641 is a perfect square. The above expression is already grouped into equal factors:

14641 = (11 x 11) x (11 x 11)

= (11 x 11)2

Hence, 14641 is the square of 121, which is equal to 11 x 11.

(iv) 4761 = 3 x 3 x 23 x 23

Grouping the factors into pairs of equal factors, we obtain:

4761 = (3 x 3) x (23 x 23)

No factors are left over. Hence, 4761 is a perfect square. The above expression is already grouped into equal factors:

4761 = (3 x 23) x (3 x 23)

= (3 x 23)2

Hence, 4761 is the square of 69, which is equal to 3 x 23.