How many pairs $(m, n)$ exist?

Question

For certain pairs $ (m,n)$ of positive integers with $ m\ge n$ there are exactly $ 50$ distinct positive integers $ k$ such that $ |\log m - \log k| < \log n$. Find the sum of all possible values of the product $ mn$.

HINTS ONLY!

Obviously, converting it into a simpler form is good idea.

$$\frac{1}{n} < \frac{m}{k} < n \implies k < mn < n^2k.$$

Okay, so we already have a pretty simplified form, I broke it into two cases.

Case 1: $m = n$ then saw:

$\implies k < n^2 < kn^2$, which is true for ALL $k \ge 2$ and any $n$.

So $m= n$ is an impossible case.

The remaining case left is $m > n$.

This is actually getting quite difficult.

There must be exactly $50$ values of $k$.

How should I go about this?

SMALL - HINTS ONLY! Please don't give it away!

Answer 1

HINT :

We have $$\frac 1n\lt\frac mk\lt n\iff \frac mn\lt k\lt mn\tag 1$$

Setting $\lfloor\frac mn\rfloor=s$ gives $$(1)\iff s+1\le k\le mn-1$$with $s\le\frac mn\lt s+1$.

Hence, one has $50=(mn-1)-(s+1)+1$.

Answer 2

Perhaps it will help to see

$$|\log m-\log k| < \log n \implies \log m-\log n < \log k < \log m+\log n \implies \log\frac{m}{n} < \log k < \log mn$$

Where $\log$ is monotonic.