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

Mock AIME 2 Pre 2005 (Problems, Source)
Preceded by Problem 4	Followed by Problem 6
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Mock AIME 2 Pre 2005 Problems/Problem 5

Problem

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