IMO 1976 SL 11

Origin: VIE

Problem

Prove that there exist infinitely many positive integers n such that the decimal representation of 5n contains a block of 1976 consecutive zeros.

Solution

We shall show by induction that 52k −1 = 2k+2qk for each k = 0, 1, . . . , where qk \inN. Indeed, the statement is true for k = 0, and if it holds for some k then 52k+1 −1 =  52k + 1  52k −1

= 2k+3dk+1 where dk+1 =  52k + 1

dk/2 is an integer by the inductive hypothesis. Let us now choose n = 2k + k + 2. We have 5n = 10k+2qk + 5k+2. It follows from 54 < 103 that 5k+2 has at most [3(k+2)/4]+2 nonzero digits, while 10k+2qk ends in k + 2 zeros. Hence the decimal representation of 5n contains at least [(k + 2)/4] −2 consecutive zeros. Now it suffices to take k > 4 \cdot 1978.