Solution:
Such a number cannot be divisible by $2$ (or its last digit would be even) or by $5$ (or its last digit would be $0$ or $5$). So if the result is false then the number must be of the form $3^{m} 7^{n}$ for non-negative integers $m, n$. But we claim that a number of this form must have even $10$s digit.

It is easy to prove the claim by induction. It is true for $3$ and $7$ (the digit is $0$ in both cases). But if we multiply such a number by $3$ or $7$, then the new $10$s digit has the same parity as the carry from the units digit. But multiplying $1,3,7,9$ by $3$ gives a carry of $0,0,2,6$ respectively, which is always even, and multiplying by $7$ gives a carry of $0,2,4,6$, which is also always even. So the new number also has an even $10$s digit.