jmc

Question

Let S be the set of integers n > 1 for which 1n = 0.d_1d_2d_3d_4, 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.)

Accepted Answer

Let k = d_1 d_2 d_3 d_12, the first 12 decimal digits of 1n. We can see that(10^12 - 1)(1n) = k kn = 10^12 - 1,so S is the set that contains all divisors of 10^12 - 1 except for 1. Since10^12 - 1 = (10^6 + 1)(10^6 - 1) = (10^2 + 1)(10^4 - 10^2 + 1)(10^3 + 1)(10^3 - 1) = 101 9901 37 11 13 7 3^3 37,the number 10^12 -1 has 4 2^6 = 256 divisors and our answer is 256 - 1 = 255.