Chapter 1 Solutions — A Square and a Cube

A perfect square ends only in 0, 1, 4, 5, 6, or 9 in the units place. Using prime factorisation to confirm:

(i) 2032 = 2⁴ × 127. The factor 127 appears only once (odd exponent), so the prime factors cannot be split into two equal groups. 2032 is NOT a perfect square.

(ii) 2048 = 2¹¹. The exponent 11 is odd, so the factors cannot be split into two equal groups. 2048 is NOT a perfect square.

(iii) 1027 = 13 × 79. Both primes appear only once (odd exponents), so 1027 is NOT a perfect square.

(iv) 1089 = 3² × 11². All prime factors have even exponents, so 1089 = (3 × 11)² = 33². 1089 IS a perfect square.

Therefore, (i) 2032, (ii) 2048, and (iii) 1027 are not perfect squares.

Method: prime factorise, then split factors into three identical groups.

Cube root of 27000:

27000 = 27 × 1000 = 3³ × 10³ = 3³ × (2 × 5)³ = 2³ × 3³ × 5³.

Group into three identical sets: (2 × 3 × 5) × (2 × 3 × 5) × (2 × 3 × 5) = 30³.

Therefore ³√27000 = 30.

Verify: 30³ = 30 × 30 × 30 = 27000 ✓.

Cube root of 10648:

10648 ÷ 2 = 5324, ÷ 2 = 2662, ÷ 2 = 1331.

1331 ÷ 11 = 121, ÷ 11 = 11, ÷ 11 = 1.

10648 = 2³ × 11³.

Group into three identical sets: (2 × 11) × (2 × 11) × (2 × 11) = 22³.

Therefore ³√10648 = 22.

Verify: 22³ = 22 × 22 × 22 = 10648 ✓.

The last digit of a square depends only on the last digit of the original number.

64² : last digit of 64 is 4. 4 × 4 = 16, so last digit of 64² is 6.

108²: last digit of 108 is 8. 8 × 8 = 64, so last digit of 108² is 4.

292²: last digit of 292 is 2. 2 × 2 = 4, so last digit of 292² is 4.

36² : last digit of 36 is 6. 6 × 6 = 36, so last digit of 36² is 6.

Verify: 108² = 11664 (last digit 4 ✓), 292² = 85264 (last digit 4 ✓).