Determining Whether the Number $11111$ is Prime. Used Divisibility Tests.

Best I can do:

Let $p$ be a prime dividing $11111$. Then I claim that $p\equiv 1 \pmod 5$

Pf: Indeed, $11111=\frac 19\times (10^5-1)$ so $p\,|\,11111\implies p\,|\,10^5-1$ which implies that $10$ has order $5\pmod p$. Thus $5\,|\,p-1$ and we are done.

Thus you should just check $11,31,41\cdots$ and stop since $41$ already works.

Here is a list for test of prime factor of less than $50$.

Test for divisibility by $41$. Subtract four times the last digit from the remaining leading truncated number. If the result is divisible by $41$, then so was the first number. Apply this rule over and over again as necessary.

$$1111-4(1)=1107$$ $$110-4(7)=110-28=82$$ $$8-4(2)=0$$

The number is divisible by $41$.

Not a clean method though but I used Fermat's factorization to find that,

$(105)^2<11111<(106)^2$

Now applying the fact that a perfect square should end only in $0,1,4,5,6,9$, concentrate only on those numbers, the difference of whose square and $11111$ give these digits in the last place. For instance, omit $113,123,133,143$ because the squares of these numbers end with $9$ and would result in $8$ as the last digit and thus this difference will not be a perfect square.

Similarly omit numbers like $112, 122, 132, 142, 152$ (Can you see the reason why?)

On omitting more of such numbers and a little bit of trial, we find that $(156)^2 - 11111 = 13225=(115)^2$

Therefore, $11111 = (156-115)(156+115) = 41.271$