Question

The numbers 528 and 540, written as the products

of their prime factors, are 528 = 24 x 3 × 11 and
540=22 x 3³ x 5. Hence, find
(i) the smallest non-zero whole number h for
which 528h is a multiple of 540,
(ii) the smallest whole number k for which
a factor of 540

Answer 1

Answer: To find the smallest non-zero whole number h for which 528h is a multiple of 540, we need to find the smallest positive whole number h such that 528h is divisible by 540.

First, we can simplify the prime factorization of 540:

540 = 22 x 3³ x 5

To make 528h a multiple of 540, we need to ensure that the prime factors of 528h are the same as the prime factors of 540.

528 = 24 x 3 x 11

So, 528h = (24 x 3 x 11)h = 24h x 3h x 11h

We can see that 24h and 11h are not factors of 540.

So, the only way to make 528h a multiple of 540 is to make 3h a factor of 3³.

3h = 3³ = 27

So, h = 27/3 = 9

So, the smallest non-zero whole number h for which 528h is a multiple of 540 is 9.

For (ii)

The smallest whole number k for which a factor of 540

540 = 22 x 3³ x 5

The smallest whole number k that is a factor of 540 is 5.

Explanation: