Question

(2) Why do the following numbers are not perfect square?
(i) 1057 (m) 7928
(m) 222
(iv) 640​

Answer 1

The prime factorization of 222 = 21 × 31 × 371. Here, the prime factor 2 is not in the pair. Therefore, 222 is not a perfect square.

Because in a square no there are even number of zeros but here 640 have 1 zero which is not even so this number can't be a square.

1057 is not a perfect square. Since, the ending digit is 7 (which is not one of 0, 1, 4, 5, 6 or 9). ... Since, the ending digit is 8 (which is not one of 0, 1, 4, 5, 6 or 9). ∴ 7928 is not a perfect square.

Answer 2

We know that, numbers having 2, 3, 7 or 8 at units place are not perfect squares.

That is because:

1

2

=1

2

2

=4

3

2

=9

4

2

=16

5

2

=25

6

2

=36

7

2

=49

8

2

=64

9

2

=81

Also,

10

2

=100

100

2

=10000

Hence, any perfect square will end with 1, 4, 5, 6, 9 or even number of zeros.

All the given numbers end with either 2, 3, 7, 8 or with odd number of zeros.

Hence, the given numbers are not perfect squares