TITLE: USAMO problem hint.
QUESTION [6 upvotes]: Prove that for every positive integer n there exists an n-digit number divisible by 5$^n$ all of whose digits are odd.
USAMO 2003.
This is the first time i have seen a problem like this, so i am not sure what to do, induction, construction, checking small cases, contradiction are some of the things i have tried.
I know I can easily find a solution anywhere but i don't want to look at a solution so please give HINTS.
I HAVE POSTED A SOLUTION USAMO problem solution HERE, PLEASE DO CHECK IT OUT.
Please don't give the full solution, any hints would be appreciated.

REPLY [6 votes]: Hint: following lulu's comment, let's suppose you have formed a number $N$ with $n-1$ odd digits divisible by $5^{n-1}$. Let's write this number as $N = p\cdot5^{n-1}$. Then you want to find an odd digit $a$ such that $a\cdot10^{n-1}+ p\cdot5^{n-1} = k\cdot5^n$ for some integer $k > 0$. This is true iff $5 | (a\cdot2^{n-1}+p)$. Writing $a = 2m+1$, can you prove that we can always find $m$? Also $m$ is mod $5$, and hence $a$ is a digit.
The base case is obvious.