How many positive multiples of 2013, and have exactly 2013 divisors?

Question

The full question:

How many positive integers are multiples of 2013 and have exactly 2013 divisors (including 1 and the number itself)?

I thought of 2013!, but I think it wouldn't work out well, since it would have more than 2013 divisors. I don't have any other way to go through this problem, so it would be greatly appreciated if anyone can answer this with an explanation...

score 2 · Answer 1 · answered Mar 19 '21 at 07:35

2

$2013 = 3 \times 11 \times 61$.

Now, the number of divisors of $n=\prod p_i^{\alpha_i}$ is $\prod(\alpha_i+1)$.

Clearly, $\prod(\alpha+1)=2013 \implies n=p_1^2 p_2^{10} p_3^{60}$ for some $p_1,p_2,p_3$. ALso, since $2013|n$, $\{p_1,p_2,p_3\}=\{3,11,61\}$.

So, the number of such integers is $3!=6$.

answered Mar 19 '21 at 07:35

MathIsNice1729

4,510

Thanks for the answer, though I have to ask, what symbol is that? I'm not familiar on what that does – Pete Justin Dagaraga Mar 19 '21 at 08:36
Which one? $\prod$? It's just product. It gives you the product of the arguments. In this case, it gives you the product of the $p_i$s. – MathIsNice1729 Mar 19 '21 at 08:39
Ah ok, thanks for that – Pete Justin Dagaraga Mar 19 '21 at 13:46
Since you are new and might not be familiar with the site, you should click the tick beside my answer and mark it as accepted if it adequately answers your question. If not, please feel free to suggest corrections and/or ask for clarification. – MathIsNice1729 Mar 19 '21 at 14:19

How many positive multiples of 2013, and have exactly 2013 divisors?

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