Question

By which smallest number must the following numbers be divided so that the quotient is a perfect cube?

(i) 675 

(ii) 8640

(iii) 1600 

(iv) 8788

(v) 7803

Answer 1

(i) 675

675 = 3 × 3 × 3 × 5 × 5

= 3³ × 5²

Since 675 is not a perfect cube.

To make the quotient a perfect cube we divide it by 5² = 25, which gives 27 as quotient where, 27 is a perfect cube.

∴ 25 is the required smallest number.

(ii) 8640

8640 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 5

= 2³ × 2³ × 3³ × 5

Since 8640 is not a perfect cube.

To make the quotient a perfect cube we divide it by 5, which gives 1728 as quotient and we know that 1728 is a perfect cube.

∴5 is the required smallest number.

(iii) 1600

1600 = 2 × 2 × 2 × 2 × 2 × 2 × 5 × 5

= 2³ × 2³ × 5²

Since 1600 is not a perfect cube.

To make the quotient a perfect cube we divide it by 5² = 25, which gives 64 as quotient and we know that 64 is a perfect cube

∴ 25 is the required smallest number.

(iv) 8788

8788 = 2 × 2 × 13 × 13 × 13

= 2² × 13³

Since 8788 is not a perfect cube.

To make the quotient a perfect cube we divide it by 4, which gives 2197 as quotient and we know that 2197 is a perfect cube

∴ 4 is the required smallest number.

(v) 7803

7803 = 3 × 3 × 3 × 17 × 17

= 3³ × 17²

Since 7803 is not a perfect cube.

To make the quotient a perfect cube we divide it by 17² = 289, which gives 27 as quotient and we know that 27 is a perfect cube

∴ 289 is the required smallest number