Express 1400 as a product of its prime factors. Answer ... q is a number between 70 and 90.
The highest common factor of q and 1400 is 28.
Find the smallest possible value of q. Answer q = ...

1. Decompose 1400 into prime factors and write the product in index form.

$1400 = 14 \times 100$
$= (2 \times 7)(2 \times 2 \times 5 \times 5)$
$= 2^3 \times 5^2 \times 7$

1. Use the HCF condition to identify possible multiples of 28 between 70 and 90.

From part (a),
$1400 = 2^3 \times 5^2 \times 7$.

Since $\text{HCF}(q,1400)=28$ and $28=2^2\times 7$, $q$ must be a multiple of $28$.

Multiples of $28$ between $70$ and $90$:
$28\times 3=84$
So the only possible value is $q=84$.

2. Check that the HCF is exactly 28.

Check:
$84=2^2\times 3\times 7$

Common prime factors of $84$ and $1400$ are
$2^2\times 7=28.$

Thus $\text{HCF}(84,1400)=28$.