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We can find number with consists only from ones and zeros. We will first watch number A=100..0100...100..1.....100..01 with k ones and between two consecutive ones exactly $ \phi(n)-1$ zeros. If this number is divisible by n then OK. Else change position of some one by moving it for a place to the left and some one not move more than once( if it was 1000001 then it is possible to get numbers 1000010 and 10000001). Obviously in begging A was congruent with k mod n. After first move it is congruent with k+9, after second k+18 etc. Because $k>=n$ and $n$ is not divisible by 3 so will reach that remainder be 0. If n isn't coprime with 10 it isn't big problem.

Using the idea to construct $m$ with only $1$'s and $0$'s, we want to show there exists distinct nonnegative integers $a_1,...a_k$ with $N$ divides $10^{a_1}+....+10^{a_k}$. First assume $N$ is coprime to $10$. Then we can find infinitely many $a$ with $10^a=1 \mod N$ and $10^a=10 \mod N$ (by choose $\phi(n)$ divides $a$ and $\phi(n)$ divides $a-1$, respectiviely). We will make $j$ of the $a_i$'s equal to $10 \mod N$ and the other $k-j$ equal to $1 \mod N$, where $0 \le j \le k$. We need $N$ divides $10j+(k-j)=k+9j$. Because $N$ is relatively prime to $3$, there exists a solution $j$ to $9j=-k \mod N$ with $ 0 \le j <N \le k$. Note that by the above construction, we can make the $a_i$'s arbitrarily large, so we can get $m$ divisible by arbitrarily large powers of $10$. This covers the case when $N$ is not coprime to $10$.

!_! 6 year period revives

Factors of $2$ and $5$ in $n$ do not matter since we can always add zeroes to the end of our number to make divisibility true. Thus, it suffices to show the same result when $\gcd(n,30)=1$. In fact, we will show it when the only digits present are zeroes and ones. Consider the number \[10^{a_1}+10^{a_2}+10^{a_3}+\dots +10^{a_k}\]where $a_1> a_2> \dots >a_k$. Since $10^{a_1}$ is periodic, if $S_n$ is the set of possible remainders that $10^{a}$ can have upon division by $n$, then it remains to show that for any $k\ge n$, there exist $k$ not necessarily distinct values from $S_n$ that sum to $0\pmod n$. We start with them all being ones. Then, we can increase the sum by $9$ by changing a one to a ten. We can do this $k$ times. Since $\gcd(n,3)=1$ and $k\ge n$, \[\{k,k+9,\dots ,k+9k\}\]must contain a number divisible by $n$. We are done.

Claim: There exists such an $m$ for all $n$ relatively prime to $10$ and $3$. Proof: Since $10^x$ is periodic mod $n$, it suffices to show there exist positive integers $a_1, a_2, \ldots, a_k$ such that $10^{a_1} + 10^{a_2} + \cdots + 10^{a_k} \equiv 0\pmod n$. If we choose all $a_i = \phi(n)$ so that $10^{a_i} \equiv 1\pmod n$ and increase some of the $a_i$'s by exactly one, we get a possible sum of $k + 9l$ for any $0\le l \le k$. Choose $l$ such that $9l \equiv -k \pmod n$. Since $3\nmid n$, there exists some $l\le n \le k$ satisfying this, as desired. $\square$ If $2\mid n$ or $5\mid n$, let $b = \frac{n}{ 10^{\max(\nu_2(n), \nu_5(n))}}$ and consider the construction for $b$ (since $k \ge n \ge b$) and then add at least $\max(\nu_2(n), \nu_5(n))$ zeroes.

Assume that $n$ is relatively prime to $10$; if not, we can reduce to a number relatively prime to $10$ by adding trailing zeros to $m$. Now we assume $m$ only has the digits $0$ and $1$ and is therefore expressed as a sum of distinct powers of $10$. Furthermore assume each power of $10$ is either $1$ or $10$ modulo $n$, with $a$ of the former and $b$ of the latter. Thus $a+b=k$. It would suffice to show that there exist $a$ and $b$ so that $n\mid a+10b=9b+k$. As $b\in [0,k]$ we can use $k\ge n$ and $\gcd(n,9)=1$ to finish. In particular choose $b\in [0,n)$ such that \[b\equiv -k\cdot 9^{-1}\pmod n.\]$\blacksquare$