Difference between revisions of "2012 AIME II Problems/Problem 11"
Williamhu888 (talk | contribs) (Created page with "== Problem 11 == Let <math>f_1(x) = \frac23 - \frac3{3x+1}</math>, and for <math>n \ge 2</math>, define <math>f_n(x) = f_1(f_{n-1}(x))</math>. The value of <math>x</math> that sa...") |
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== Problem 11 == | == Problem 11 == | ||
Let <math>f_1(x) = \frac23 - \frac3{3x+1}</math>, and for <math>n \ge 2</math>, define <math>f_n(x) = f_1(f_{n-1}(x))</math>. The value of <math>x</math> that satisfies <math>f_{1001}(x) = x-3</math> can be expressed in the form <math>\frac mn</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>. | Let <math>f_1(x) = \frac23 - \frac3{3x+1}</math>, and for <math>n \ge 2</math>, define <math>f_n(x) = f_1(f_{n-1}(x))</math>. The value of <math>x</math> that satisfies <math>f_{1001}(x) = x-3</math> can be expressed in the form <math>\frac mn</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>. | ||
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| + | == Solution == | ||
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| + | == See also == | ||
| + | {{AIME box|year=2012|n=II|num-b=10|num-a=12}} | ||
Revision as of 17:21, 31 March 2012
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
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 | ||
