Difference between revisions of "2012 AIME II Problems/Problem 11"
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== Solution == | == Solution == | ||
| − | After evaluating the first few values of <math>f_k (x)</math>, we obtain <math>f_4(x) = f_1(x) = \frac{2}{3} - \frac{3}{3x+1} = \frac{6x-7}{9x+3}</math>. Since <math>1001 \equiv 2 | + | After evaluating the first few values of <math>f_k (x)</math>, we obtain <math>f_4(x) = f_1(x) = \frac{2}{3} - \frac{3}{3x+1} = \frac{6x-7}{9x+3}</math>. Since <math>1001 \equiv 2 \mod 3</math>, <math>f_{1001}(x) = f_2(x) = \frac{3x+7}{6-9x}</math>. We set this equal to <math>x-3</math>, i.e. |
| − | <math>\frac{3x+7}{6-9x} = x-3 \Rightarrow x = \frac{5}{3}</math>. The answer is 5+3 = 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> |
== 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}} | ||
Revision as of 18:57, 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
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 | ||
