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+{\Huge Proof of a Riddle}

+{\LARGE Written by: Shlomi Fish}

+We need to prove that for every natural number $n > 0$, there exists a

+decimal number of $n$ digits, which can be wholly divided by $5^n$, and

+all of its digits are odd.

+We will prove a stronger claim. We will demonstrate that if $b_n$ is the

+corresponding number for $n$, then it can serve as a suffix for $b_{n+1}$

+, by adding another most significant digit.

+\item For every $n$, there exists an $a \in \{1,3,5,7,9\}$ so that

+$b_{n+1} = b_n + a \cdot 10^n$ and $b_{n+1} \mod 5^{n+1} = 0 $.

+The proof would be by induction.

+\subsection{Induction Base}

+It holds for $n = 1$ as 5 is a one-digit number that is wholly divisable

+\subsection{Induction Step}

+Let's assume it holds for $n$ and show that it also holds for $n+1$.

+\[b_{n+1} = b_n + a \cdot 10^n \]

+According to the induction step $b_n$ is wholly divisable by $5^n$ and so

+is $10^n = 5^n \cdot 2^n$. So we can divide the expression by $5^n$ and

+try to find an $a$ so that the quotient is divisable by 5. We get:

+\[ b_{n}' + a \cdot 2^n \]

+$b_{n}'$ has some modulo 5, and $2^n$ has a non-zero modulo. The values that

+$a$ can assume (1,3,5,7,9) contain all the modulos of 5. Since 5 is prime

+, and its modulos are a group, we can get all modulos by multiplying a

+given non-zero modulo by the other modulos. So we can choose an $a$ so that

+the expression modulo 5 evaluates to 0. Thus we can divide this $b_{n+1}$ by