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Difference between revisions of "1984 AIME Problems/Problem 7"

(→‎Solution 2)
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We start by finding values of the function right under 1000 since they require little repetition.
 
We start by finding values of the function right under 1000 since they require little repetition.
  
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<cmath>\begin{align*}f(999)=f(f(1004))=f(1001)=998\\
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<math>f(999)=f(f(1004))=f(1001)=998</math>
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&f(998)=f(f(1003))=f(1000)=997</cmath>
+
<math>f(998)=f(f(1003))=f(1000)=997</math>
−
f(997)=f(f(1002))=f(999)=998
+
<math>f(997)=f(f(1002))=f(999)=998</math>
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f(996)=f(f(1001))=f(998)=997\\
+
<math>f(996)=f(f(1001))=f(998)=997</math>
  
 
Soon we realize the <math>f(k)</math> for integers <math>k<1000</math> either equal <math>998</math> or <math>997</math> based on it parity. (If short on time, a guess of 998 or 997 can be taken now.)
 
Soon we realize the <math>f(k)</math> for integers <math>k<1000</math> either equal <math>998</math> or <math>997</math> based on it parity. (If short on time, a guess of 998 or 997 can be taken now.)

Revision as of 17:31, 25 July 2011

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 $f(1004) = 999$ is the answer; however, that is not true since the solution occurs slightly before that. Start at $f^3(1004)$: \[f^{3}(1004)=f^{2}(1001)=f(998)=f^{2}(1003)=f(1000)=\boxed{997}\]


Solution 2

We start by finding values of the function right under 1000 since they require little repetition.

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

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 • Answer Key • Resources)
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
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