Question

For each integer $n \ge 4,$ let $a_n$ denote the base- $n$ number $0.\overline{133}_n.$ The product $a_4 a_5 \ldots a_{99}$ can be expressed as $\dfrac{m}{n!},$ where $m$ and $n$ are positive integers and $n$ is as small as possible. What is the value of $m?$

$98$

$101$

$132$

$798$

$962$

Solution:

Since $n^3 \cdot a_n = 133.\overline{133}_n = a_n + n^2 + 3n + 3,$ we get $a_n = \dfrac{n^2 + 3n + 3}{n^3 - 1} = \dfrac{(n+1)^3 - 1}{n(n^3 - 1)}.$

Writing $n^3 - 1 = (n - 1)(n^2 + n + 1)$ and $(n+1)^3 - 1 = n\big((n+1)^2 + (n+1) + 1\big),$ the product $a_4 a_5 \cdots a_{99}$ telescopes to $\dfrac{3!}{99!} \cdot \dfrac{100^3 - 1}{6^3} = \dfrac{3!}{99!} \cdot \dfrac{99(100^2 + 100 + 1)}{63}.$

This simplifies to $\dfrac{(2)(10101)}{(21)(98!)} = \dfrac{962}{98!},$ so $m = 962$ (with the smallest possible $n = 98$ ).

Thus, the correct answer is E.

2004 AMC 12A Problem 25

Solution:

Problem 25 in Other Years