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>\frac{3x+7}{6-9x} = x-3 \Rightarrow x = \frac{5}{3}</math>. The answer is thus <math>5+3 = \boxed{008}</math>. |
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+ | ==Video Solution== | ||
+ | https://www.youtube.com/watch?v=zBKm3M71K4c&t=47s | ||
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+ | This video is now private. | ||
== 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}} |
Latest revision as of 17:03, 29 December 2023
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.youtube.com/watch?v=zBKm3M71K4c&t=47s
This video is now private.
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.