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Difference between revisions of "2009 AIME II Problems/Problem 11"

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Remember that we must have <math>m\geq n</math>. However, for <math>n\geq 8</math> we have <math>\dfrac{51n}{n^2-1} < n</math>, and hence <math>m<n</math>, which is a contradiction.
 
Remember that we must have <math>m\geq n</math>. However, for <math>n\geq 8</math> we have <math>\dfrac{51n}{n^2-1} < n</math>, and hence <math>m<n</math>, which is a contradiction.
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This only leaves us with the cases <math>n\in\{2,3,4,5,6\}</math>.
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This only leaves us with the cases <math>n\in\{2,3,4,5,6,7\}</math>.
  
 
* For <math>n=2</math> we have <math>\dfrac{100}3 < m \leq \dfrac{102}3</math> with a single integer solution <math>m=\dfrac{102}3=34</math>.
 
* For <math>n=2</math> we have <math>\dfrac{100}3 < m \leq \dfrac{102}3</math> with a single integer solution <math>m=\dfrac{102}3=34</math>.
 
* For <math>n=3</math> we have <math>\dfrac{150}8 < m \leq \dfrac{153}8</math> with a single integer solution <math>m=\dfrac{152}8=19</math>.
 
* For <math>n=3</math> we have <math>\dfrac{150}8 < m \leq \dfrac{153}8</math> with a single integer solution <math>m=\dfrac{152}8=19</math>.
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* For <math>n=4,5,6</math> our inequality has no integer solutions for <math>m</math>.
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* For <math>n=4,5,6,7</math> our inequality has no integer solutions for <math>m</math>.
  
 
Therefore the answer is <math>34\cdot 2 + 19\cdot 3 = 68 + 57 = \boxed{125}</math>.
 
Therefore the answer is <math>34\cdot 2 + 19\cdot 3 = 68 + 57 = \boxed{125}</math>.
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{{AIME box|year=2009|n=II|num-b=10|num-a=12}}
 
{{AIME box|year=2009|n=II|num-b=10|num-a=12}}
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[[Category: Intermediate Algebra Problems]]
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{{MAA Notice}}

Revision as of 15:18, 21 May 2023

Problem

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 $mn$.

Solution

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 $n\in\{2,3,4,5,6,7\}$.

  • 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 $n=4,5,6,7$ 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 (Problems • Answer Key • Resources)
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

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