Written as a product of its prime factors, $p = 2^1 × 3^x × 7^y$ and $q = 3 × 7^2 × 11$

1. Write 21 as a product of prime factors.

$21=3\times 7$
For $p$ to be a multiple of 21, $p$ must contain at least one factor of $3$ and at least one factor of $7$.

2. Choose the smallest exponents of 3 and 7 in $p$.

Since
$p=2^1\times 3^x\times 7^y,$
the smallest powers needed are
$3^1 \text{ and } 7^1.$
Hence,
$x=1,\quad y=1.$

1. Write $33q$ in prime factorised form.

Since $q=3\times7^2\times11$ and $33=3\times11$,
$33q=(3\times11)(3\times7^2\times11)=3^2\times7^2\times11^2.$

2. Explain why the result is a perfect square.

In $3^2\times7^2\times11^2$, every prime factor has an even exponent. Therefore
$33q=(3\times7\times11)^2,$
so $33q$ is a perfect square.