been having headache trying to understand this question. Can I clarify if the question is asking the largest factor that when multiples with n^3 will be 24? I don’t understand the relationship between n must have primes in multiples of 3 but then the answer becomes n must contain at least one 2 and one 3.

Try choosing numbers.

Tried tackling this with a refreshed mind. Let me know if my working is correct

By prime factorising 24, I gets 2^3.3. I have to imagine n as two separate numbers. I could get 24 with the following combo

2^3 x 3

2^2 x 2 x 3 = 2^2 x 6

2 x 2^2 x 3 = 2^2 x 6

The only largest factor made possible is 6.

Is this reasoning correct?

Btw when you mentioned try choosing numbers, how would you choose the number in this scenario?

I don’t quite understand what you’re trying to say, especially this part.

Try looking for an *easy* number. So n = 6 for instance. That automatically short-circuits the answer.

I mentioned imagining two numbers because none of the answer choices if you take a cube will get to 24.

We take 6 because the answer has to be a factor to another cubic number whose multiplication has to be divisible by 24, is that correct? My apologies if I made it even more convoluted.

OK I think I get what you meant.

But is that reasoning correct? Thank you in advance and for your patience

That is 2^3 \times 3, not 2^2 \times 6? I don’t see a cubic multiplied by 6.