# 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 $n+1$ times. Then we have $$f(84)=f^{n+1}(84)=f(f^n(84+5))$$ In order to find $f(84)$, we want to know the smallest value of $$f^n(84+5)\ge1000$$ Because then $$f(84)=f(f^n(84+5))=(f^n(84+5))-3$$ From which we'll get a numerical value for $f(84)$.

Notice that the value of $n$ we expect to find is basically the smallest $n$ such that after $f(x)=f(f(x+5))$ is performed $\frac{n}{2}$ times and then $f(x)=x-3$ is performed back $\frac{n}{2}$ times, the result is greater than or equal to $1000$.

In this case, the value of $n$ for $f(84)$ is $916$, because $$84+\frac{916}{2}\cdot5-\frac{916}{2}\cdot3=1000\Longrightarrow f^{916}(84+5))=1000$$ Thus $$f(84)=f(f^{916}(84+5))=f(1000)=1000-3=\boxed{997}$$

~ Nafer