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.
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 $19^2*53= 19133$.