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. | ||
− | This only leaves us with the cases <math>n\in\{2,3,4,5,6\}</math>. | + | 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>. | ||
− | * For <math>n=4,5,6</math> our inequality has no integer solutions for <math>m</math>. | + | * 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}} | ||
+ | [[Category: Intermediate Algebra Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 14:18, 21 May 2023
Problem
For certain pairs of positive integers with there are exactly distinct positive integers such that . Find the sum of all possible values of the product .
Solution
We have , hence we can rewrite the inequality as follows: We can now get rid of the logarithms, obtaining: And this can be rewritten in terms of as
From it follows that the solutions for must be the integers . This will happen if and only if the lower bound on is in a suitable range -- we must have .
Obviously there is no solution for . For the left inequality can be rewritten as , and the right one as .
Remember that we must have . However, for we have , and hence , which is a contradiction. This only leaves us with the cases .
- For we have with a single integer solution .
- For we have with a single integer solution .
- For our inequality has no integer solutions for .
Therefore the answer is .
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.