MathDB

Problem 5

Part of 2011 CIIM

Problems(1)

CIIM 2011 Problem 5

Source:

6/9/2016
Let nn be a positive integer with dd digits, all different from zero. For k=0,...,d−1k = 0,. . . , d - 1, we define nkn_k as the number obtained by moving the last kk digits of nn to the beginning. For example, if n=2184n = 2184 then n0=2184,n1=4218,n2=8421,n3=1842n_0 = 2184, n_1 = 4218, n_2 = 8421, n_3 = 1842. For mm a positive integer, define sm(n)s_m(n) as the number of values kk ​​such that nkn_k is a multiple of m.m. Finally, define ada_d as the number of integers nn with dd digits all nonzero, for which s2(n)+s3(n)+s5(n)=2d.s_2 (n) + s_3 (n) + s_5 (n) = 2d. Find lim⁡d→∞ad5d.\lim_{d \to \infty} \frac{a_d}{5^d}.
CIIM 2011undergraduateCIIM