2012 AIME II Problems/Problem 11: Difference between revisions
mNo edit summary |
|||
| (6 intermediate revisions by 4 users not shown) | |||
| Line 7: | Line 7: | ||
<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>. | ||
==Video Solution== | |||
https://www.youtube.com/watch?v=zBKm3M71K4c&t=47s | |||
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}} | |||
Latest revision as of 16:03, 29 December 2023
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 | ||
These problems are copyrighted © by the Mathematical Association of America.