# 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 $a_4a_5\cdots a_{99}$ can be expressed as $\frac {m}{n!}$, where $m$ and $n$ are positive integers and $n$ is as small as possible. What is $m$? $\text {(A)} 98 \qquad \text {(B)} 101 \qquad \text {(C)} 132\qquad \text {(D)} 798\qquad \text {(E)}962$

## 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{align*} 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{align*} \begin{align*} a_4a_5\cdots 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\cdots 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{align*}

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

## Solution 2

Note that $$0.\overline{133}_n = \frac{n^2+3n+3}{n^3-1},$$by geometric series. Thus, we're aiming to find the value of $$\prod_{k=4}^{99} \frac{k^2+3k+3}{k^3 - 1}.$$Expanding the product out, this is equivalent to $$\frac{4^2+3(4)+3}{4^3 - 1} \cdot \frac{5^2+3(5)+3}{5^3 - 1} \cdot \frac{6^2+3(6)+3}{6^3 - 1} \cdot ... \cdot \frac{99^2+3(99)+3}{99^3 - 1}.$$Note that the numerator of the $a$th fraction and the denominator of the $a+1$th fraction for $1 \leq a \leq 95$ cancel out to be $\frac{1}{a+3},$ since $$\frac{k^2 + 3k + 3}{(k+1)^3 - 1} = \frac{k^2 + 3k + 3}{k^3 + 3k^2 + 3k} = \frac{1}{k},$$by the binomial theorem on the the denominator of the aforementioned. Since this forms a telescoping series, our product is now equivalent to $$\frac{99^2 + 3(99) + 3}{4^3 - 1} \cdot \prod_{k=4}^{98} \frac{1}{k},$$which, after simplification gives $\frac{6}{98!} \cdot \frac{10101}{63} = \frac{962}{98!},$ giving an answer of $\boxed{962}.$

-fidgetboss_4000

## See Also

 2004 AMC 12A (Problems • Answer Key • Resources) Preceded byProblem 24 Followed byLast Question 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. Invalid username
Login to AoPS