# Mock AIME 2 Pre 2005 Problems/Problem 5

## Problem

Let $S$ be the set of integers $n > 1$ for which $\tfrac1n = 0.d_1d_2d_3d_4\ldots$, an infinite decimal that has the property that $d_i = d_{i+12}$ for all positive integers $i$. Given that $9901$ is prime, how many positive integers are in $S$? (The $d_i$ are digits.)

## Solution

Let $k = d_1 d_2 d_3 \ldots d_{12}$, the first $12$ decimal digits of $\tfrac{1}{n}$. We can see that $$(10^{12} - 1)\left(\dfrac{1}{n}\right) = k \implies kn = 10^{12} - 1,$$ so $S$ is the set that contains all divisors of $10^{12} - 1$ except for $1$. Since $$10^{12} - 1 = (10^6 + 1)(10^6 - 1) = (10^2 + 1)(10^4 - 10^2 + 1)(10^3 + 1)(10^3 - 1) = 101 \cdot 9901 \cdot 37 \cdot 11 \cdot 13 \cdot 7 \cdot 3^3 \cdot 37,$$ the number $10^{12} -1$ has $4 \cdot 2^6 = 256$ divisors and our answer is $256 - 1 = \boxed{255}.$

-MP8148

## See also

 Mock AIME 2 Pre 2005 (Problems, Source) Preceded byProblem 4 Followed byProblem 6 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15