Difference between revisions of "2012 AIME II Problems/Problem 11"
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<math>\frac{3x+7}{6-9x} = x-3 \Rightarrow x = \frac{5}{3}</math>. The answer is thus <math>5+3 = \boxed{008}</math>. | <math>\frac{3x+7}{6-9x} = x-3 \Rightarrow x = \frac{5}{3}</math>. The answer is thus <math>5+3 = \boxed{008}</math>. | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://www.youtu.be/zBKm3M71K4c&t=47s | ||
== See Also == | == See Also == | ||
{{AIME box|year=2012|n=II|num-b=10|num-a=12}} | {{AIME box|year=2012|n=II|num-b=10|num-a=12}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 14:24, 8 July 2020
Contents
Problem 11
Let , and for , define . The value of that satisfies can be expressed in the form , where and are relatively prime positive integers. Find .
Solution
After evaluating the first few values of , we obtain . Since , . We set this equal to , i.e.
. The answer is thus .
Video Solution
https://www.youtu.be/zBKm3M71K4c&t=47s
See Also
2012 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.