# For every positive number $n$, there exists a $n$ digit number having all odd digits and divisible by $5^n$

Prove that for every positive integer $n$, there exist a $n$ digit number, divisible by $5^n$, whose all digits are odd. for example,

for $n=1, 5$

$n=2,75$

$n=3, 375$.......

• I don't have an answer yet, but I notice that the solutions for increasing $n$ are all the same, except that each one has an additional digit on the front. For example, you have 5, then 75, then 375, and the next one is 9​​375. So there's probably some way to say that if $s_n$ is an $n$-digit solution, then you can choose $k$ from $1,3,5,7,9$ so that $k\cdot10^{n-1} + s_n$ is an $n+1$-digit solution. You already know that it has all odd digits, so the only matter is showing that it is divisible by $5^n$, and you automatically have that it is divisible by $5^{n-1}$. – MJD Jun 28 '14 at 13:48
We prove this by induction. You have already shown that it is true for $n=1,2,3$.
Suppose that for some specific $n\ge3$ we have an integer $a=5^nb$ such that $a$ has $n$ digits and every digit is odd. Consider the number $a'=a+10^nk$, where $1\le k<10$. Clearly this is a number of $n+1$ digits; all of them, except perhaps the first, are odd; we shall show that it is possible to choose $k$, also odd, such that $a'$ is a multiple of $5^{n+1}$.
To make $a'$ a multiple of $5^{n+1}$ we require $5^{n+1}\mid 5^nb+10^nk$, that is, $$5\mid b+2^nk\ .$$ Since $5$ and $2^n$ are coprime, this has a unique solution with $1\le k\le5$. If $k=2$ or $k=4$ we can replace it by $k=7$ or $k=9$ respectively; therefore we can always satisfy the divisibility condition with an odd value of $k$, and the proof is complete.