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1984 AIME Problems/Problem 7

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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^{... 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.

See also

1984 AIME (Problems • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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