Part a is simply "divide until you cannot divide".

1350
= 135 x 10
= (27 x 5) x (2 x 5)
= (9 x 3 x 5) x (2 x 5)
= (3 x 3 x 3 x 5) x (2 x 5)
= 2 x 3³ x 5²

So, we have

720 = 2^4 x 3² x 5
1350 = 2 x 3³ x 5²

For the HCF, we pick the lower power available for each prime factor.

We pick one 2 (because the lower power of 2 is 1), two 3s (because the lower power of 3 is 2) and one 5 (because the lower power of 5 is 1).

So, the HCF
= 2 x 3² x 5
= 90

For the LCM, we pick the higher power available for each prime factor.

We pick four 2s, three 3s and two 5s.

So, the LCM
= 2^4 x 3³ x 5²
= 10800

if there is
1) 2^4 x 3^2 x 5 x 7 and 2) 2 x 3^3 x 5^2

will the HCF be 2 x 3^3 x 5^2 as there is no 7 in (2) which means 7^0 is the lowest power

and LCM is 2^4 x 3^3 x 5^2 x 7?

btw can you help me for part d as well... dont really know how to start the qn

If one of the numbers lacks a factor which the other number possesses,

- the HCF will not contain that factor,
- the LCM will contain that factor and with the highest power possible

720 = 2^4 x 3² x 5

For 720k to be a perfect cube, you need each of the powers to be a multiple of 3.

(This is because we need to distribute the factors into three equal groups so that picking one group is essentially the same as taking cube roots)

So, we need two more 2s, one more 3 and two more 5s, all coming from the k.

k
= 2² x 3 x 5²
= 300

thx :)