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

Revision as of 18:26, 10 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 $f^3(1004)$ :

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

@@ Line 1: / Line 1: @@
-==Problem==
+== Problem ==
-A finite [[sequence]] of three-digit integers has the property that the tens and units digits of each term are, respectively, the hundreds and tens digits of the next term, and the tens and units digits of the last term are, respectively, the hundreds and tens digits of the first term. For example, such a sequence might begin with the terms 247, 475, and 756 and end with the term 824. Let <math>S</math> be the sum of all the terms in the sequence. What is the largest [[prime]] [[factor]] that always divides <math>S</math>?
+The [[function]] f is defined on the [[set]] of [[integer]]s and satisfies
+<math>
+f(n)=
+\begin{cases}
+ n-3 & \mbox{if }n\ge 1000 \\
+ f(f(n+5)) & \mbox{if }n<1000
+\end{cases}
+</math>
-<math>\mathrm{(A)}\ 3\qquad \mathrm{(B)}\ 7\qquad \mathrm{(C)}\ 13\qquad \mathrm{(D)}\ 37\qquad \mathrm{(E)}\ 43</math>
+Find <math>\displaystyle f(84)</math>.
-==Solution==
+== Solution ==
-A given digit appears as the hundreds digit, the tens digit, and the units digit of a term the same number of times. Let <math>k</math> be the sum of the units digits in all the terms. Then <math>S=111k=3*37k</math>, so <math>S</math> must be divisible by <math>37\ \mathrm{(D)}</math>. To see that it need not be divisible by any larger prime, the sequence <math>123, 231, 312</math> gives <math>S=666=2*9*37</math>.
+Define <math>\displaystyle f^{h}(x) = 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>.
-==See also==
+Let’s write out a couple more iterations of this function:
-{{AMC12 box|year=2007|ab=A|num-b=10|num-a=12}}
-[[Category:Introductory Algebra Problems]]
+<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>
+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>:
+<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 ==
+{{AIME box|year=1984|num-b=6|num-a=8}}
+* [[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 18:26, 10 September 2007

Problem

Solution

See also