# The following numbers are obviously not perfect squares. Give reason

(i) 1057 (ii) 23453 (iii) 7928 (iv) 222222 (v) 64000

(vi) 89722 (vii) 222000 (viii) 505050

**Solution:**

The square of a number having 0, 1, 4, 5, 6 or 9 at its unit place is perfect squares. Also, the square of a number can only have an even number of zeros at the end.

In the above question unit digit of numbers (i) 1057 (ii) 23453 (iii) 7928 (iv) 222222 (v) 64000

(vi) 89722 (vii) 222000 (viii) 505050 are 7, 3, 8, 2, 000, 2, 000, 0 respectively.

So these numbers are obviously not perfect squares as they do not end with the digits 0, 1, 4, 5, 6 or 9 and also have odd number of zeros.

**☛ Check: **NCERT Solutions for Class 8 Maths Chapter 6

**Video Solution:**

## The following numbers are obviously not perfect squares. Give reason. (i) 1057 (ii) 23453 (iii) 7928 (iv) 222222 (v) 64000 (vi) 89722 (vii) 222000 (viii) 505050

NCERT Solutions for Class 8 Maths Chapter 6 Exercise 6.1 Question 2

**Summary**:

The following numbers are obviously not perfect squares (i) 1057 (ii) 23453 (iii) 7928 (iv) 222222 (v) 64000 (vi) 89722 (vii) 222000 (viii) 505050 because they neither end with 0, 1, 4, 5, 6, 9 nor end with an even number of zeroes

**☛ Related Questions:**

- The squares of which of the following would be odd numbers? (i) 431 (ii) 2826 (iii) 7779 (iv) 82004
- Observe the following pattern and find the missing digits. 112 = 121 1012 = 10201 10012 = 1002001 1000012 = 1....2....1 100000012 = ...........
- Observe the following pattern and supply the missing numbers. 112 = 121 1012 = 10201 101012 = 102030201 10101012 = ? ?2 = 10203040504030201
- Using the given pattern, find the missing numbers. 12 + 22 + 22 = 32 22 + 32 + 62 = 72 32 + 42 + 122 = 132 42 + 52 + _2 = 212 52 + -2 + 302 = 312 62 + 72 + _2 = _2

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