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

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== Problem ==
 
== Problem ==
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The [[function]] f is defined on the [[set]] of [[integer]]s and satisfies
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The [[function]] f is defined on the [[set]] of [[integer]]s and satisfies <math>f(n)=\begin{cases}
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<math>
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n-3&\mbox{if}\ n\ge 1000\\
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f(n)=
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f(f(n+5))&\mbox{if}\ n<1000\end{cases}</math>
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\begin{cases}
 
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n-3 & \mbox{if }n\ge 1000 \\
 
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f(f(n+5)) & \mbox{if }n<1000
 
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\end{cases}
 
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</math>
 
  
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Find <math>\displaystyle f(84)</math>.
+
Find <math>f(84)</math>.
  
 
== Solution ==
 
== Solution ==
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Define <math>\displaystyle f^{h} = f(f(\cdots f(f(x))\cdots))</math>, where the function <math>f</math> is performed <math>h</math> times. We find that <math> f(84) = f(f(89) = f^2(89) = f^3(94) = \ldots f^{y}(1004)</math>. <math>\displaystyle 1004 = 84 + 5(y - 1) \Longrightarrow y = 185</math>. So we now need to reduce <math>\displaystyle f^{185}(1004)</math>.
+
Define <math>f^{h} = f(f(\cdots f(f(x))\cdots))</math>, where the function <math>f</math> is performed <math>h</math> times. We find that <math> f(84) = f(f(89) = f^2(89) = f^3(94) = \ldots f^{y}(1004)</math>. <math>1004 = 84 + 5(y - 1) \Longrightarrow y = 185</math>. So we now need to reduce <math>f^{185}(1004)</math>.
  
 
Let’s write out a couple more iterations of this function:
 
Let’s write out a couple more iterations of this function:
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<cmath>\begin{align*}f^{185}(1004)&=f^{184}(1001)=f^{183}(998)=f^{184}(1003)=f^{183}(1000)\\
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<div style="text-align:center;"><math>\displaystyle f^{185}(1004) = f^{184}(1001) = f^{183}(998)</math><br /><math>= f^{184}(1003) = f^{183}(1000) = f^{182}(997)</math><br /><math>= f^{183}(1002) = f^{182}(999) = f^{183}(1004)</math></div>
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&=f^{182}(997)=f^{183}(1002)=f^{182}(999)=f^{183}(1004)\end{align*}</cmath>
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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) = 999</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>
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<div style="text-align:center;"><math>f^{3}(1004) = f^{2}(1001) = f(998)</math><br /><math>= f^{2}(1003) = f(1000) = 997</math></div>
 
  
 
== See also ==
 
== See also ==

Revision as of 19:21, 25 April 2008

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

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 $f^3(1004)$: \[f^{3}(1004)=f^{2}(1001)=f(998)=f^{2}(1003)=f(1000)=\boxed{997}\]

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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