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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.

Aditya_math
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    My first thought would be to proceed by induction. Assume you've done it for $n-1$ and then argue that you can put an odd digit in front of your number to handle $n$. Haven't tried it so I'm not certain in works. Good luck! – lulu Nov 22 '20 at 17:26
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    Hint: every $5^n$ such that $n \geq 2$ has an even digit $2$. Try ways to eliminate it (example, $5^2 = 25$, $25 \times 3 = 75$ all of whose digits are odd). There may also occur $6$s as well. See if you could eliminate them... – Spectre Nov 22 '20 at 17:38
  • @lulu i think i have solved it using induction, can i post my solution to check for any mistakes – Aditya_math Nov 22 '20 at 18:40
  • Sure. I'll look it over once you've posted it – lulu Nov 22 '20 at 18:40
  • @lulu i have posted a solution https://math.stackexchange.com/questions/3918561/usamo-problem-solution – Aditya_math Nov 22 '20 at 19:20
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    I think your solution is fine. It's not necessary to check each case explicitly, though there is no harm in doing so. Indeed, having those case by case solutions helps if you are writing code to list the examples. – lulu Nov 22 '20 at 19:38

1 Answers1

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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.

Peanut
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