CIIM 2011 Problem 5

Let $n$ be a positive integer with $d$ digits, all different from zero. For $k = 0,. . . , d - 1$ , we define $n_k$ as the number obtained by moving the last $k$ digits of $n$ to the beginning. For example, if $n = 2184$ then $n_0 = 2184, n_1 = 4218, n_2 = 8421, n_3 = 1842$ . For $m$ a positive integer, define $s_m(n)$ as the number of values $k$ such that $n_k$ is a multiple of $m.$ Finally, define $a_d$ as the number of integers $n$ with $d$ digits all nonzero, for which $s_2 (n) + s_3 (n) + s_5 (n) = 2d.$ Find $$\lim_{d \to \infty} \frac{a_d}{5^d}.$$