For what values of $n$ (positive integer) does 19 divide $10^n-1$ evenly? The question arises in my fooling around with $2$-parasitic numbers. And I know this is true for $n = 18, 36, 54$ (by computing!) I expect this to be so for all multiples of $18.$ Would appreciate any feedback.
Thanks in advance.
That is correct. Once you know that $19\mid 10^{18}-1$ you can note that $10^{18k}-1=(10^{18}-1)(1+10^{18}+10^{36}+\cdots+ 10^{18(k-1)})$
It is indeed.
A proof relies on lil' Fermat: we know that $\varphi(19)=18$, hence for all $N$ coprime to $19$, we have $N^18\equiv 1$\mod 19. This is true for $10$, which means its order modulo $19$ is a divisor of $18$, i.e. one of $2,3,6, 9$ or $18$. It is straightforward to check recursively it is $18$.
To answer your question as to whether $10^k \equiv_1 19$ only for $k$ that is a multiple of 18: You can easily check the following: $10^{18+\ell} \equiv_{19} 10^{\ell}$ for each nonegative integer $\ell$. Indeed,
$$10^{18+\ell} \mod 19 = ((10^{18} \mod 19)\times(10^{\ell} \mod 19) \mod 19)$$
$$=1 \times (10^{\ell} \mod 19)$$
So calculate $10^{\ell} \mod 19$ for each $\ell=2,3,4,5,6,\ldots, 17$. This can be done via brute force without too much trouble.
