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
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 $50=(mn-1)-(s+1)+1$.
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