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

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(→‎Solution 1)
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<cmath>\begin{align*}f^{185}(1004)&=f^{184}(1001)=f^{183}(998)=f^{184}(1003)=f^{183}(1000)\\
 
<cmath>\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*}</cmath>
 
&=f^{182}(997)=f^{183}(1002)=f^{182}(999)=f^{183}(1004)\end{align*}</cmath>
−
So this function reiterates with a period of 2 for <math>x</math>. It might be tempting at first to assume that <math>f(1004) = 999</math> is the answer; however, that is not true since the solution occurs slightly before that. Start at <math>f^3(1004)</math>:
+
So this function reiterates with a period of 2 for <math>x</math>. It might be tempting at first to assume that <math>f(1004) = 1001</math> is the answer; however, that is not true since the solution occurs slightly before that. Start at <math>f^3(1004)</math>:
 
<cmath>f^{3}(1004)=f^{2}(1001)=f(998)=f^{2}(1003)=f(1000)=\boxed{997}</cmath>
 
<cmath>f^{3}(1004)=f^{2}(1001)=f(998)=f^{2}(1003)=f(1000)=\boxed{997}</cmath>
−
 
  
 
== Solution 2==
 
== Solution 2==

Revision as of 10:09, 5 April 2012

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 $f(84)=997$.

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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