Prove that for every odd integer n, exist a multiple m of n whose decimal representation entirely consists of odd digits.

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Korbin Ochoa · Accepted Answer

Step 1First I claim there is an n digit number whose digits are odd and which is divisible by 5n. It’s true for n=1. Assume it’s true for n−1. That number is u×5n−1 for some integer u≡0,1,2,3,4bmod5 Now 10n−1 is also v×5n−1 for some v≡1,2,3,4bmod5,, so by adding a first digit which is 1,3,5,7,9 the new number is the same power of 5 timesu+v, u+3v, u+5v, u+7v, u+9v bmod5,And exactly one of these will be divisible by 5, and so one is done.This produces a number m with n digits divisible by 5n. The number mmmmmmmmm… Is thenm×10kn−110n−1And then taking kn divisible by φ((10n−1)r) we can for (r,10)=1 we can get divisiblity by 5n×r by a number with only odd digits.

Prove that for every odd integer n, exist

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