2004 AMC 12A Problem 25

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

A repeating base-

n

fraction

0.\overline{133}_n

equals

\dfrac{n^2 + 3n + 3}{n^3 - 1}

Note

n^2 + 3n + 3 = \dfrac{(n+1)^3 - 1}{n},

which makes the product telescope

Since

n^3 \cdot a_n = 133.\overline{133}_n

= a_n + n^2 + 3n + 3,

we get

\begin{aligned} a_n &= \dfrac{n^2 + 3n + 3}{n^3 - 1} \\ &= \dfrac{(n+1)^3 - 1}{n(n^3 - 1)}. \end{aligned}

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

\begin{gathered} \dfrac{3!}{99!} \cdot \dfrac{100^3 - 1}{4^3 - 1} \\ {}= \dfrac{3!}{99!} \cdot \dfrac{99(100^2 + 100 + 1)}{63}. \end{gathered}

This simplifies to

\dfrac{(2)(10101)}{(21)(98!)} = \dfrac{962}{98!}.

If

n \le 97,

then rewriting this fraction with denominator

n!

would require

98

to divide

962,

which it does not. Hence the smallest possible

n

is

98,

and

m = 962.

Thus, the correct answer is E.

Problem 25 in other years