Decimal expansion of $kn$ contains only $0,7$ for some $k \in \mathbb{N}$

Question

Show that for any positive integer $n$, there exists a positive multiple of $n$ that contains only the digits $0$ and $7$ in its decimal expansion.

I don't know how to tackle this problem, a hint would be nice…

Answer 1

It is sufficient to show that we can construct some number $r$ with only $0$, $7$ in its decimal expansion such that it is divisible by some set $p_1^{n_1},...,p_k^{n_k}$ (the $p_i$ are primes) for arbitrary choices of $p_i$, $k_i$, since if $n=p_1^{n_1}...p_k^{n_k}$ then $n$ divides $r$. Let the decimal expansion of this number have $7$s at positions $a_1,...,a_l$ (so if we have $a_1=1$, $a_2=4$ then the number is $7007$).

For $p_i \neq 7$ (for $7$ we're fine anyway) we want $r=\sum_{j=1}^{l}7^{a_j}=0 \pmod{p_i^{n_i}}$. Letting $i$ run over the different primes in our set we receive the system of simultaneous equations:

$\sum_{j=1}^{l}7^{a_j}=0 \pmod{p_1^{n_1}}$

...

$\sum_{j=1}^{l}7^{a_j}=0 \pmod{p_k^{n_k}}$.

Using the fact that $7^{\phi(p_i^{k_i})}=1 \pmod{p_i^{k_i}}$ I think you should be able to construct something from here (hint: can you find a number $x$ such that $7^x=1\pmod{p_i^{k_i}} \ \ \forall i$? hint2: now you have $x$, can you find another one? hint3: so what to do with them?)