Does every number not ending with zero have a multiple without zero digits at all?

Solution 1:

Then answer is yes.

If $n$ is relatively prime to $10$ then the answer is easy, see for example this post.

If, $n=2^k m$ with $m$ relatively prime to $10$ or $n=5^km$ with $m$ relatively prime to $10$, then prove first the following by induction:

Claim 1 $2^k$ has a $k$ digits multiple containing only the digits $1$ and $2$.

Claim 2 $5^k$ has a $k$ digits multiple containing only the digits $1,2,3,4$ and $5$.

Both are easy induction exercises.

Once you prove this, proceed as follows (same idea as in the posted link):

$n=2^k m$ or $n=5^km$ with $gcd(m,10)=1$.. Then, we know that $2^k$ respectively $5^k$ has a multiple of the form $\overline{a_1...a_k}$

Look at the following numbers $$\overline{a_1...a_k} , \overline{a_1...a_ka_1...a_k} , \overline{a_1...a_ka_1...a_ka_1...a_k} , ...$$

In this infinite list, there exists two numbers with the same reminder when divided by $m$. Their difference is a multiple of $m$.

Therefore $$m| \overline{a_1...a_ka_1...a_ka_1...a_k00...0}=\overline{a_1...a_ka_1...a_ka_1...a_k}\cdot 10^l$$ Since $m$ is relatively prime to $10$ it follows that $$m|\overline{a_1...a_ka_1...a_ka_1...a_k}$$ Since $2^k$ respectively $5^k$ also divide $\overline{a_1...a_ka_1...a_ka_1...a_k}$, and they are relatively prime to $m$, it follows that $$n|\overline{a_1...a_ka_1...a_ka_1...a_k}$$

P.S. We proved that $n$ has a multiple which can be written only with the digits $1,2,3,4,5$. These digits can be replaced by any $5$ digits which cover all classes $\pmod{2}$ and $\pmod{5}$.