1984 AIME Problems/Problem 7

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 $f(84)$ .

Solution 1

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

Let’s write out a couple more iterations of this function: $\begin{align*}f^{185}(1004)&=f^{184}(1001)=f^{183}(998)=f^{184}(1003)=f^{183}(1000)\\ &=f^{182}(997)=f^{183}(1002)=f^{182}(999)=f^{183}(1004)\end{align*}$ So this function reiterates with a period of 2 for $x$ . It might be tempting at first to assume that $f(1004) = 1001$ is the answer; however, that is not true since the solution occurs slightly before that. Start at $f^3(1004)$ : $f^{3}(1004)=f^{2}(1001)=f(998)=f^{2}(1003)=f(1000)=\boxed{997}$

Solution 2

We start by finding values of the function right under $1000$ since they require iteration of the function.

$f(999)=f(f(1004))=f(1001)=998$ $f(998)=f(f(1003))=f(1000)=997$ $f(997)=f(f(1002))=f(999)=998$ $f(996)=f(f(1001))=f(998)=997$

Soon we realize the $f(k)$ for integers $k<1000$ either equal $998$ or $997$ based on it parity. (If short on time, a guess of $998$ or $997$ can be taken now.) If $k$ is even $f(k)=997$ if $k$ is odd $f(k)=998$ . $84$ has even parity, so $f(84)=997$ . The result may be rigorously shown through induction.

Solution 3

Assume that $f(84)$ is to be performed $n$ times for some natural number $n$ . Then we have $f^n(84)=f(f^{n-1}(89)$

1984 AIME (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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1984 AIME Problems/Problem 7

Contents

Problem

Solution 1

Solution 2

Solution 3

See also