Let integer $n>10$ such that $n=\overline{a_{m}a_{m-1}a_{m-2}...a_{0}}$ where: $a_i \in {(1;3;7;9)}$, $i=\overline{0,m}$ Prove that: $n$ has a prime divisor which is not smaller than $11$.
Here or what I tried:
Notice that: $n$ is odd
Now $n$ can be written as:
$$n=(10^{a_1}+10^{a_2}+...10^{a_n}).1+(10^{b_1}+10^{b_2}+...10^{b_n}).3+(10^{c_1}+10^{c_2}+...10^{c_n}).7+(10^{d_1}+10^{d_2}+...10^{d_n}).9$$
Have: $$10 \equiv -1 mod 11$$
But frome there, I can't make anything more. Can anyone continue with this please, my teacher said that I should use number system like decimal system!