2007 AMC 12A Problems/Problem 11

Revision as of 18:26, 10 September 2007 by Azjps (talk | contribs) (solution)

Problem

The function f is defined on the set of integers and satisfies $f(n)= \begin{cases}  n-3 & \mbox{if }n\ge 1000 \\  f(f(n+5)) & \mbox{if }n<1000 \end{cases}$

Find $\displaystyle f(84)$.

Solution

Define $\displaystyle f^{h}(x) = f(f(\cdots f(f(x))\cdots))$, where the function $f$ is performed $h$ times. We find that $f(84) = f(f(89) = f^2(89) = f^3(94) = \ldots f^{y}(1004)$. $\displaystyle 1004 = 84 + 5(y - 1) \Longrightarrow y = 185$. So we now need to reduce $\displaystyle f^{185}(1004)$.

Let’s write out a couple more iterations of this function:

$\displaystyle f^{185}(1004) = f^{184}(1001) = f^{183}(998)$
$= f^{184}(1003) = f^{183}(1000) = f^{182}(997)$
$= f^{183}(1002) = f^{182}(999) = f^{183}(1004)$

So this function reiterates with a period of 2 for $x$. It might be tempting at first to assume that $f(1004) = 999$ is the answer; however, that is not true since the solution occurs slightly before that. Start at $f^3(1004)$:

$f^{3}(1004) = f^{2}(1001) = f(998)$
$= f^{2}(1003) = f(1000) = 997$

See also

1984 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions