For every integer, some multiple of it is of the form $99 \ldots 900 \ldots 00$

Question

The goal is to prove that for every positive integer $ z$ there exists a positive integer $a$ such that $az = 99 \ldots 9900 \ldots 00$.

Let $a = \frac {99 \ldots 9900 \ldots 00}{z}$

That doesn't look right. How do I fix it?

Answer 1

Suppose we have a number $k$, lets say it has $k_0$ trailing zeroes and $k_d$ digits that are not trailing zeroes. The trick is to multiply $k$ by $2$ until the number formed by the $k_d$ non-trailing zeroes is odd and does not end in $5$. note that the number formed by $k_d$(we reffer to it as $K_d$) is relatively prime to $10$. Therefore there is a power of $10$ which is congruent to 1 mod $K_d$. So we have a solution to $10^n-1\equiv0 \bmod K^d$. So then it is possible to multiply $k$ by $2$ repeatedly until $K_d$ ends with an odd number that is not $5$. And once this happens we can make $K_d$ of the form $999\dots999$ and once that happens we are done because the trailing zeroes will stay unchanged. .

Answer 2

Consider the numbers $u_n=99\cdots9$, where $u_n$ has $n$ nines.

Step 1: They are not all different mod $a$.

Step 2: if $u_n$ and $u_m$ agree mod $a$, with $n>m$, then $u_n-u_m$ is your solution.