Express 1008 as a product of prime factors, leaving your answer in index notation. Answer 1008 = ... p and q are numbers such that q is a prime number. Find the values of p and q such that $\dfrac{1008p}{q}$ is the smallest possible cube. Answer p = ..., q = ... The lowest common multiple of x and 1008 is $2^{5}\times3^{2}\times5\times7$. Find the smallest possible value of x. Answer x = ...

1. Find the prime factorisation of 1008 and express repeated factors using indices.

$1008=2\times504$

$=2\times2\times252$

$=2\times2\times2\times126$

$=2\times2\times2\times2\times63$

$=2^4\times3\times3\times7$

$=2^4\times3^2\times7$

1. Use the prime factorisation of 1008 and test which prime q gives the smallest cube multiplier.

From part (a)(i),
$1008=2^4\times3^2\times7.$
Since $q$ is prime and $\dfrac{1008p}{q}$ must be an integer cube, $q$ should be one of the prime factors of $1008$: $2$, $3$ or $7$.

To make the result the smallest possible cube, divide by $7$:
$\frac{1008}{7}=2^4\times3^2.$

2. Find the smallest p that makes all exponents multiples of 3.

For a cube, each exponent must be a multiple of $3$.
$2^4\times3^2\times p$
needs one more factor of $2^2$ and one more factor of $3$:
$p=2^2\times3=12.$
Then
$\frac{1008\times12}{7}=2^6\times3^3,$
which is a cube.

1. Compare the prime powers in $1008$ with the required LCM and choose the extra prime powers needed for the smallest $x$.

From part (a)(i),
$1008=2^4\times3^2\times7.$
The required LCM is
$2^5\times3^2\times5\times7.$
For $\operatorname{lcm}(x,1008)$ to have $2^5$, $x$ must contain $2^5$.
For it to contain $5$, $x$ must contain $5$.
The factors $3^2$ and $7$ are already supplied by $1008$, so they need not be included in the smallest $x$.
$x=2^5\times5=32\times5=160.$