2009 AIME II Problems/Problem 11


For certain pairs $(m,n)$ of positive integers with $m\geq n$ 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 $mn$.


We have $\log m - \log k = \log \left( \frac mk \right)$, hence we can rewrite the inequality as follows: \[- \log n < \log \left( \frac mk \right) < \log n\] We can now get rid of the logarithms, obtaining: \[\frac 1n < \frac mk < n\] And this can be rewritten in terms of $k$ as \[\frac mn < k < mn\]

From $k<mn$ it follows that the $50$ solutions for $k$ must be the integers $mn-1, mn-2, \dots, mn-50$. This will happen if and only if the lower bound on $k$ is in a suitable range -- we must have $mn-51 \leq \frac mn < mn-50$.

Obviously there is no solution for $n=1$. For $n>1$ the left inequality can be rewritten as $m\leq\dfrac{51n}{n^2-1}$, and the right one as $m > \dfrac{50n}{n^2-1}$.

Remember that we must have $m\geq n$. However, for $n\geq 8$ we have $\dfrac{51n}{n^2-1} < n$, and hence $m<n$, which is a contradiction. This only leaves us with the cases $n\in\{2,3,4,5,6\}$.

  • For $n=2$ we have $\dfrac{100}3 < m \leq \dfrac{102}3$ with a single integer solution $m=\dfrac{102}3=34$.
  • For $n=3$ we have $\dfrac{150}8 < m \leq \dfrac{153}8$ with a single integer solution $m=\dfrac{152}8=19$.
  • For $n=4,5,6$ our inequality has no integer solutions for $m$.

Therefore the answer is $34\cdot 2 + 19\cdot 3 = 68 + 57 = \boxed{125}$.

See Also

2009 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS