Every prime number has multiple of form $100\dots01$

Question

Does Every Prime number (except $2,3,5$) has a multiple of form $10^k+1$ where $k\in N$?

I check till $29$ and it seems to be true. However it got gigantic. $29\times3448275862069=10^{14}+1$

I Proved using pigeonhole that every prime (except $2,5$) has multiple of form $10^k-1$ but am not able to do it with $10^k+1$

Answer 1

$$10^3-1=999=27\times 37.$$ So $10^3\equiv1\pmod {37}$. Modulo $37$ the powers of ten are $1,10,26,1,10,26,1,\ldots$. None of these are $\equiv-1\pmod{37}$.

Answer 2

Note: $10^k\equiv -1 \pmod p\implies$ the order of $10\pmod p$ is even (if $L$ is the least exponent such that $10^L\equiv -1\pmod p$ then the order of $10\pmod p$ is $2L$).

The order of $10\pmod {31}$ is $15$ which is odd, so the claim fails in that case.