The proof is by induction. The property is obviously true for n = 1: 5 (a single odd digit number) is divisible by 51.
Assume that N = a1a2...an = 5n ·M, M not divisible by 5 and all ai's are odd. Consider the numbers
N1 = 1a1a2...an = 1·10n + 5n ·M = 5n(1·2n + M),
N3 = 3a1a2...an = 3·10n + 5n ·M = 5n(3·2n + M),
N5 = 5a1a2...an = 5·10n + 5n ·M = 5n(5·2n + M),
N7 = 7a1a2...an = 7·10n + 5n ·M = 5n(7·2n + M),
N9 = 9a1a2...an = 9·10n + 5n ·M = 5n(9·2n + M).
The multiples 1·2n + M, 3·2n + M, ..., 9·2n + M all have different remainders when divided by 5. (Otherwise, the difference of some two of them would be divisible by 5, which is obviously not true.) It follows that one of them is divisible by 5; the corresponding number N is then divisible by 5n·5 = 5n+1.
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