It is a question I always ask when I was a primary student:
Could a number only contains digit'1' with $p$ digits writes $$A=\frac{10^{p}-1}{9} = \underbrace{11\cdots 1}_{p\text{ digits}}$$ be a prime?
When $p$ is not a prime, the factorization of $A$ is obvious, but if $p$ is a prime, the factorization of $A$ may be subtle.
Here I made a list of some result for prime $p$
$$ \begin{array}{c|lcr} p & \ \text{factors} \\ \hline 3 & 3 \times 37\\ 5 & 41 \times 271 \\ 7 & 239 \times 4649 \\ 11 & 21649 \times 513239 \\ 13 & 53 \times 79 \times 265371653 \\ 17 & 2071723 \times 5363222357 \\ 19 & \text{prime} \\ 23 & \text{prime} \\ 29 & 3191 \times 16763 \times 43037 \times 62003 \times 77843839397 \\ 31 & 2791 \times 6943319 \times 57336415063790604359 \\ 37 & 2028119 \times 247629013 \times 2212394296770203368013 \\ 41 & 83 \times 1231 \times 538987 \times 201763709900322803748657942361 \\ 43 & 173 \times 1527791 \times 1963506722254397 \times 2140992015395526641 \\ 47 & 35121409 \times 316362908763458525001406154038726382279 \\ 53 & 107 \times 1659431 \times 1325815267337711173 \times 47198858799491425660200071 \\ 59 & 2559647034361 \times 4340876285657460212144534289928559826755746751 \\ 61 & 733 \times 4637 \times 329401 \times 974293 \times 1360682471 \times 106007173861643 \times 7061709990156159479 \\ 67 & 493121 \times 79863595778924342083 \times 28213380943176667001263153660999177245677 \\ \end{array} $$
I am not major in number theory, just curious again if there is any theorem or former work on this kind of question.