Express 396 as the product of its prime factors. Answer ___ Given that $16\,200 = 2^3 \times 3^4 \times 5^2$, find the smallest possible integer value of $k$ such that $396k$ is a multiple of $16\,200$. Answer k = ___ the smallest possible integer value of $p$ such that $\dfrac{16\,200}{p}$ is a cube number. Answer p = ___

1. Divide 396 into prime factors and express using powers.

$396=2\times 198=2\times 2\times 99=2^2\times 9\times 11=2^2\times 3^2\times 11$

1. Compare prime factors and find the missing factors needed in $k$.

From part (a),
$396=2^2\times 3^2\times 11$.

Given
$16\,200=2^3\times 3^4\times 5^2$.

For $396k$ to be a multiple of $16\,200$, $k$ must supply the missing factors:
$k=2^{3-2}\times 3^{4-2}\times 5^2$
$k=2\times 3^2\times 5^2=2\times 9\times 25=450$

1. Compare the prime exponents with multiples of 3 to make the quotient a cube number.

Given
$16200=2^3\times 3^4\times 5^2$.

For $\dfrac{16200}{p}$ to be a cube number, each remaining prime exponent must be a multiple of $3$.

Keep the largest cube factor in $16200$:
$2^3\times 3^3\times 5^0$
So divide away the extra factors:
$p=3^{4-3}\times 5^2=3\times25=75$