1.9 Finding the Smallest k for Perfect Squares and Cubes

Given a number, find the smallest positive $k$ making the product, or the quotient, a perfect square or cube. Factorise, then fix whichever indices do not yet qualify.

Write the number in index notation.
For a square, every index must be even; for a cube, a multiple of $3$ .
MULTIPLYING by $k$ : $k$ supplies whatever each short index still needs.
DIVIDING by $k$ : $k$ removes the surplus instead.

Finding smallest k for a perfect cube

Given that $600 = 2^3 \times 3 \times 5^2$ , find the smallest positive integer $k$ such that $600k$ is a perfect cube.

$600 = 2^3 \times 3^1 \times 5^2$
For a perfect cube, each index must be a multiple of $3$ .
Index of $2$ is $3$ : already a multiple of $3$ . ✓
Index of $3$ is $1$ : need $1 \to 3$ , so multiply by $3^2$ .
Index of $5$ is $2$ : need $2 \to 3$ , so multiply by $5^1$ .
$k = 3^2 \times 5 = 45$
Check: $600 \times 45 = 27\,000 = 2^3 \times 3^3 \times 5^3 = 30^3$ ✓

Try it

A practice question on this appears here. Practise primes hcf lcm.

Dividing to make a perfect square

Given that $8232 = 2^3 \times 3 \times 7^3$ , find the smallest whole number $m$ such that $\frac{8232}{m}$ is a perfect square. Hence find $\sqrt{\frac{8232}{m}}$ .

$8232 = 2^3 \times 3^1 \times 7^3$
For a perfect square, each index must be even.
Index of $2$ is $3$ (odd): divide by $2^1$ to get $2^2$ .
Index of $3$ is $1$ (odd): divide by $3^1$ to get $3^0$ .
Index of $7$ is $3$ (odd): divide by $7^1$ to get $7^2$ .
$m = 2 \times 3 \times 7 = 42$
$\frac{8232}{42} = 196 = 2^2 \times 7^2$
$\sqrt{196} = 2 \times 7 = 14$

Practise this

Questions on Primes, HCF and LCM, marked as you go, with the working shown.

Practise Primes, HCF and LCM