2004 AMC 12A Problems/Problem 25

Problem

For each integer $n\geq 4$ , let $a_n$ denote the base- $n$ number $0.\overline{133}_n$ . The product $a_4a_5...a_{99}$ can be expressed as $\frac {m}{n!}$ , where $m$ and $n$ are positive integers and $n$ is as small as possible. What is the value of $m$ ?

$\text {(A)} 98 \qquad \text {(B)} 101 \qquad \text {(C)} 132\qquad \text {(D)} 798\qquad \text {(E)}962$

Solution

$a_x = \frac{1}{x}+\frac{3}{x^2}+\frac{3}{x^3}+\frac{1}{x^4}+\frac{3}{x^5}+\frac{3}{x^6}+\cdots$

$a_x*x^3=x^2+3x+3+a_x$

$a_x(x^3-1)=x^2+3x+3$

$a_x=\frac{x^2+3x+3}{x^3-1}=\frac{(x+1)^3-1}{x(x^3-1)}$

$a_4a_5...a_{99}=\frac{(5^3-1)(6^3-1)\cdots (100^3-1)}{4*5*6*\cdots*99*(4^3-1)(5^3-1)\cdots(99^3-1)}$

$a_4a_5...a_{99}=\frac{999999}{4*5*6*\cdots*99*63}=\frac{13*37*33*6}{99!}$

Since $13*37*33*6$ isn't one of the answer choices, we need to get rid of some stuff:

$99=33*3$

$\frac{13*37*33*6}{99!}=\frac{13*37*2}{98!}$

Since only the two goes into 98, n is at it's minimum.

$13*37*2=962 \Rightarrow \text {(E)}$

2004 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 24	Followed by Final Question
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2004 AMC 12A Problems/Problem 25

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