Difference between revisions of "2012 AIME II Problems/Problem 11"
Tempaccount (talk | contribs) (Adding problem section) |
Tempaccount (talk | contribs) (Remove extra problem section) |
||
| Line 1: | Line 1: | ||
| − | |||
| − | |||
== 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>. | ||
Revision as of 16:01, 9 August 2018
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 
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. 
