1st Balkan 1984

Prove that given any positive integer n we can find a larger integer m such that the decimal expansion of 5^m can be obtained from that for 5ⁿ by adding additional digits on the left.

Solution

We prove first that 5^N - 1 is divisible by 2ⁿ⁺² for N = 2ⁿ. This is an easy induction. It is true for n = 1. Suppose it is true for n. Then 5^N - 1 is divisible by 2ⁿ⁺². But 5^N + 1 is certainly even, so it is divisible by 2. Hence (5^N + 1)(5^N - 1) = 5^2N - 1 is divisible by 2ⁿ⁺³, so the result is true for all n.

So suppose 5ⁿ has k decimal digits. Clearly k ≤ n. We can find N such that 5^N - 1 is divisible by 2^k. Hence (5^N - 1)5ⁿ is divisible by 2^k5^k = 10^k. In other words it ends in at least k zeros. So 5^N+n = (5^N - 1)5ⁿ + 5ⁿ ends in the same digits as 5ⁿ. In other words we may take m = n + N.

1st Balkan 1984

© John Scholes
jscholes@kalva.demon.co.uk
11 Apr 2002