Compute the smallest positive integer n such that 214·n and 2014·n have the same number of divisors.

The only thing that I was able to realize in this question was the fact that this value of n needs to a divisor of either 214 or 2014. I wasn't sure how to proceed from here.

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    $\begingroup$ Have you tried? How do you compute the number of divisors of an integer in general? $\endgroup$ – Arnaud Mortier Jun 2 '18 at 2:45
  • $\begingroup$ Raise the exponent of each prime factor by 1, and then multiply all the exponents. $\endgroup$ – Dude156 Jun 2 '18 at 2:47
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    $\begingroup$ Right. If you know this it should be included in the question, as well as the prime decomposition of 214 and 2014. People are much more likely to try and help if you show that you are involved. $\endgroup$ – Arnaud Mortier Jun 2 '18 at 2:51
  • $\begingroup$ I experimented and wrote a solution @ArnaudMortier. Thanks for prompting me to try harder. $\endgroup$ – Dude156 Jun 2 '18 at 3:08

Ok I figured it out. Thanks @Arnaud Mortier for convincing me to try a bit harder. For some reason, even after realizing the crux of the problem (which was that n would have to be made up of the 2, 19, 53, 107 in some form), I still didn't give it a good effort. What I did was set $n = 2^(a-1)*19^(b-1)*53^(c-1)*107^(d-1)$. I did the -1 because I knew that I would end up raising some of these terms by +1 which would mean that I might simplify my expression. After multiplying both 214 and 2014 by this and then raising the exponents by 1, I came at the equation $(a+1)(b)(c)(d+1)=(a+1)(b+1)(c+1)d$. After some cancellations, I arrived at bc = d(b+c+1). Since I was trying to minimize this number, I set d=1 (d=0 isn't possible because we are dealing with d-1 in the exponent). This led me to get that b and c are either 2 or 3 (either order). With a goal to minimize, I set b=2 and c=3. Substituting for n and multiplying, I arrived at the answer of $19^2*53= 19133$.

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    $\begingroup$ Nice! It's not completely obvious at first that $d$ has to be $1$, but it does become clear when you see that there is a solution with $b=2$, $c=3$. A larger $d$ could not help beat that. There is a typo in the end, $b=3$ and $c=2$. $\endgroup$ – Arnaud Mortier Jun 2 '18 at 3:17

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