Express 4312 as a product of its prime factors. Given $588 = 2^2 \times 3 \times 7^2$. Find The largest integer which is a factor of both 588 and 4312. The smallest integer which is the multiple of both 588 and 4312. The smallest integer value of $m$ such that $4312m$ is a perfect square.

1. Divide by prime factors to obtain the full prime factorisation.

$4312$ is even, so divide by $2$ repeatedly:

$4312 = 2 \times 2156 = 2^2 \times 1078 = 2^3 \times 539$

Now factorise $539$:

$539 = 7 \times 77 = 7 \times 7 \times 11 = 7^2 \times 11$

Hence,

$4312 = 2^3 \times 7^2 \times 11$

1. Compare the prime factorisations and take the common prime factors with the smaller indices.

Given
$588=2^2\times3\times7^2$
and from part (a),
$4312=2^3\times7^2\times11$.

The common prime factors are $2$ and $7$.

$\text{HCF}=2^2\times7^2$

2. Evaluate the highest common factor.

$2^2\times7^2=4\times49=196$

1. Compare the prime factorisations and take each prime factor with the larger index.

From earlier parts,
$4312 = 2^3 \times 7^2 \times 11$
and
$588 = 2^2 \times 3 \times 7^2.$

Hence,
$\text{LCM} = 2^3 \times 3 \times 7^2 \times 11.$

2. Evaluate the LCM.

$2^3 \times 3 \times 7^2 \times 11 = 8 \times 3 \times 49 \times 11 = 12936.$

1. Use the prime factorisation of 4312 and identify the factors needed to make a perfect square.

From part (a),
$4312=2^3\times 7^2\times 11.$For $4312m$ to be a perfect square, all indices must be even.

$2^3$ needs one more factor of $2$, and $11^1$ needs one more factor of $11$.

So the smallest possible value is
$m=2\times 11=22.$