MathDB

CIIM 2011 Problem 5

Source:

June 9, 2016

CIIM 2011undergraduateCIIM

Problem Statement

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}.

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