Factoring numbers of the form $11111111$

Question

Why $11111111$ is divisible by $73$? How can we get all the prime factors?

It is clear that it is divisible by $11$. Is there any formulae for $1111...11$ ($n$ times)? Give me some idea. Thanks in advance.

Answer 1

$$\underbrace{11111111}_{n(\ge1) \text{ digits }}=\frac{10^n-1}9$$

Now $10^8-1=(10^4-1)(10^4+1)=(10^2-1)(10^2+1)(10^4+1)$

Clearly, $73|(10^4+1)$(why?)

I think, we don't have too many clever method other than actual division to find $10^4+1=73\cdot137$

Answer 2

The Cunningham Project has taken as its goal the factorization of lots of numbers of the form $b^n\pm1$ for $2\le b\le12$, so in particular it covers the numbers $10^n-1$ and, by extension, $(10^n-1)/9$, the number whose decimal expansion is $n$ ones. There is a lot of information at that site about the methods used, both on the theoretical side and on the computational side.