A prime number greater than 5, say P, is a factor of N, where N is 1 repeated (P - 1) times.
For e.g.:
This holds for larger numbers as well.
To check it, I used python as below:
P = 13 (any prime number)
N = '1'*(P-1)
R = N % P
Whenever P is prime(and greater than 5), R = 0. Why does this hold?
Fermat's Little Theorem provides the answer!
It states that for any prime $p$
$$a^{p-1}-1 \text{ mod(p)}\equiv 0$$
If a and p are relatively prime
Plugging in $a=10$ gives us
$$10^{p-1}-1 \text{ mod(p)} \equiv 0$$
Doing some algebra, we get
$$10^{p-1}-1 \text{ mod(p)} \equiv (10-1) * (\sum^{p-2}_{j=0}{10^j})\text{ mod(p)}\equiv0$$
And note that $$ \sum^{p-2}_{j=0}{10^j}\text{ mod(p)}\equiv0$$ is just the sum of a bunch of 1 digits, a repunit as @prets would call it.
Also, note that this is not true when $p=5$, since 10 and 5 are not relatively prime, but it is true for all prime examples greater than 5.
This is incorrect in general , take $5$ as a counter example . Notice that :
$1111= 101*11$
