Difference between revisions of "2007 AMC 12A Problems/Problem 11"

Revision as of 02:31, 11 September 2007

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 $\displaystyle f(84)$ .

Solution

Define $\displaystyle f^{h}(x) = f(f(\cdots f(f(x))\cdots))$ , where the function $f$ is performed $h$ times. We find that $f(84) = f(f(89) = f^2(89) = f^3(94) = \ldots f^{y}(1004)$ . $\displaystyle 1004 = 84 + 5(y - 1) \Longrightarrow y = 185$ . So we now need to reduce $\displaystyle f^{185}(1004)$ .

Let’s write out a couple more iterations of this function:

$\displaystyle f^{185}(1004) = f^{184}(1001) = f^{183}(998)$
$= f^{184}(1003) = f^{183}(1000) = f^{182}(997)$
$= f^{183}(1002) = f^{182}(999) = f^{183}(1004)$

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 $f^3(1004)$ :

$f^{3}(1004) = f^{2}(1001) = f(998)$
$= f^{2}(1003) = f(1000) = 997$

2007 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

@@ Line 23: / Line 23: @@
 == See also ==
-{{AIME box|year=1984|num-b=6|num-a=8}}
+{{AMC12 box|year=2007|ab=A|num-b=10|num-a=12}}
-* [[AIME Problems and Solutions]]
-* [[American Invitational Mathematics Examination]]
-* [[Mathematics competition resources]]
 [[Category:Intermediate Algebra Problems]]

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Difference between revisions of "2007 AMC 12A Problems/Problem 11"

Revision as of 02:31, 11 September 2007

Problem

Solution

See also