Difference between revisions of "2012 AIME II Problems/Problem 11"

Revision as of 15:35, 29 February 2020

Problem 11

Let $f_1(x) = \frac23 - \frac3{3x+1}$ , and for $n \ge 2$ , define $f_n(x) = f_1(f_{n-1}(x))$ . The value of $x$ that satisfies $f_{1001}(x) = x-3$ can be expressed in the form $\frac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ .

Solution

After evaluating the first few values of $f_k (x)$ , we obtain $f_4(x) = f_1(x) = \frac{2}{3} - \frac{3}{3x+1} = \frac{6x-7}{9x+3}$ . Since $1001 \equiv 2 \mod 3$ , $f_{1001}(x) = f_2(x) = \frac{3x+7}{6-9x}$ . We set this equal to $x-3$ , i.e.

$\frac{3x+7}{6-9x} = x-3 \Rightarrow x = \frac{5}{3}$ . The answer is thus $5+3 = \boxed{008}$ .

@@ 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>
+<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 ==
 {{AIME box|year=2012|n=II|num-b=10|num-a=12}}
 {{MAA Notice}}

2012 AIME II (Problems • Answer Key • Resources)
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Difference between revisions of "2012 AIME II Problems/Problem 11"

Revision as of 15:35, 29 February 2020

Problem 11

Solution

See Also