Show that for each n we can find an n-digit number with all its digits odd which is divisible by 5n.
Solution
Induction on n. For n = 1 we have 5. So suppose N works for n. Consider the five n+1 digit numbers 10n + N, 3·10n + N, 5·10n + N, 7·10n, 9·10n. We may take out the common factor 5n to get the five numbers k, k + h, k + 2h, k + 3h, k + 4h, for some k and h = 2n+1. Since h is relatively prime to 5, the five numbers are all incongruent mod 5 and so one must be a multiple of 5.
© John Scholes
jscholes@kalva.demon.co.uk
6 Aug 2003
Last corrected/updated 6 Aug 03