Problem:
Prove that for every positive integer n there exists an n-digit number divisible by 5n all of whose digits are odd.
Solution:
We proceed by induction. The property is clearly true for n=1. Assume that N=a1​a2​…an​ is divisible by 5n and has only odd digits. Consider the numbers
​N1​=1a1​a2​…an​=1⋅10n+5nM=5n(1⋅2n+M)N2​=3a1​a2​…an​=3⋅10n+5nM=5n(3⋅2n+M)N3​=5a1​a2​…an​=5⋅10n+5nM=5n(5⋅2n+M)N4​=7a1​a2​…an​=7⋅10n+5nM=5n(7⋅2n+M)N5​=9a1​a2​…an​=9⋅10n+5nM=5n(9⋅2n+M)​
The numbers 1⋅2n+M,3⋅2n+M,5⋅2n+M,7⋅2n+M,9⋅2n+M give distinct remainders when divided by 5 . Otherwise the difference of some two of them would be a multiple of 5 , which is impossible, because 2n is not a multiple of 5 , nor is the difference of any two of the numbers 1,3,5,7,9. It follows that one of the numbers N1​,N2​,N3​,N4​,N5​ is divisible by 5n⋅5, and the induction is complete.
The problems on this page are the property of the MAA's American Mathematics Competitions