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

m (→‎See also: - AMC 12 Box)
(LOL I accidentally saved an AIME q here; revert)
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== Problem ==
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==Problem==
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The [[function]] f is defined on the [[set]] of [[integer]]s and satisfies
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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>?
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<math>
 
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f(n)=
 
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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>.
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<math>\mathrm{(A)}\ 3\qquad \mathrm{(B)}\ 7\qquad \mathrm{(C)}\ 13\qquad \mathrm{(D)}\ 37\qquad \mathrm{(E)}\ 43</math>
  
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== Solution ==
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==Solution==
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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>.
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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>.
  
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Let’s write out a couple more iterations of this function:
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==See also==
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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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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>:
 
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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>
 
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== See also ==
 
 
{{AMC12 box|year=2007|ab=A|num-b=10|num-a=12}}
 
{{AMC12 box|year=2007|ab=A|num-b=10|num-a=12}}
  
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[[Category:Intermediate Algebra Problems]]
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[[Category:Introductory Algebra Problems]]

Revision as of 15:39, 11 September 2007

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 $S$ be the sum of all the terms in the sequence. What is the largest prime factor that always divides $S$?

$\mathrm{(A)}\ 3\qquad \mathrm{(B)}\ 7\qquad \mathrm{(C)}\ 13\qquad \mathrm{(D)}\ 37\qquad \mathrm{(E)}\ 43$

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 $k$ be the sum of the units digits in all the terms. Then $S=111k=3*37k$, so $S$ must be divisible by $37\ \mathrm{(D)}$. To see that it need not be divisible by any larger prime, the sequence $123, 231, 312$ gives $S=666=2*9*37$.

See also

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
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