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

This is an infinite geometric series with common ratio $\frac{1}{x^3}$ and initial term $x^{-1} + 3x^{-2} + 3x^{-3}$ , so $a_x = \left(\frac{1}{x} + \frac{3}{x^2} + \frac{3}{x^3}\right)\left(\frac{1}{1-\frac{1}{x^3}}\right)$ $= \frac{x^2 + 3x + 3}{x^3} \cdot \frac{x^3}{x^3 - 1}$ $= \frac{x^2 + 3x + 3}{x^3 - 1}$ $= \frac{(x+1)^3 - 1}{x(x^3 - 1)}$ .

Alternatively, we could have used the algebraic manipulation for repeating decimals,

$\begin{eqnarray*} a_x &=& \frac{1}{x}+\frac{3}{x^2}+\frac{3}{x^3}+\frac{1}{x^4}+\frac{3}{x^5}+\frac{3}{x^6}+\cdots \\ a_x \cdot x^3 &=& x^2+3x+3+a_x\\ a_x(x^3-1) &=& x^2+3x+3\\ a_x &=& \frac{x^2+3x+3}{x^3-1}=\frac{(x+1)^3-1}{x(x^3-1)} \end{eqnarray*}$

Telescoping,

$\begin{eqnarray*} a_4a_5...a_{99}&=&\frac{(5^3-1)(6^3-1)\cdots (100^3-1)}{4 \cdot 5 \cdot 6 \cdot \cdots \cdot 99 \cdot (4^3-1)(5^3-1)\cdots(99^3-1)}\\ a_4a_5...a_{99}&=&\frac{999999}{4 \cdot 5 \cdot 6 \cdot \cdots \cdot 99 \cdot 63}=\frac{13 \cdot 37 \cdot 33 \cdot 6}{99!}\end{eqnarray*}$

Some factors cancel, (after all, $13 \cdot 37 \cdot 33 \cdot 6$ isn't one of the answer choices)

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

Since the only factor in the numerator that goes into $98$ is $2$ , $n$ is minimized. Therefore the answer is $13 \cdot 37 \cdot 2=962 \Rightarrow \text {(E)}$ .

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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