TITLE: How many divisors does $111...1$ have?
QUESTION [5 upvotes]: Let $A=\underbrace{11..1}_{2010}$. How many divisors does $111...1$ have?
Original problem: Prove that $τ(A)>50$ (or $τ(A)<50$)

My work so far:

If $\tau(A) -$ the number of divisors of $A$ and $A=p_1^{\alpha_1}\cdotp_2^{\alpha_2}\cdot...\cdot p_n^{\alpha_n}$, then $\tau(A)=(\alpha_1+1)(\alpha_2+1)\cdot...\cdot(\alpha_n+1)$

If $a=\underbrace{11..1}_{k}$ and $b=\underbrace{11..1}_{l}$ and $l|k$ then $b|a$

$2010=2\cdot 3\cdot5\cdot67$. Then $\tau(A)\ge 2^4=16$

REPLY [3 votes]: A partial answer: note that
$$
A = \frac{\overbrace{99\cdots9}^{2010}}{9} = \frac{10^{2010} - 1}{10-1}
$$
that is, $A = \frac{x^{2010} - 1}{x-1}$ with $x = 10$.  You can exploit the many factorizations of $x^n - 1$, since $n$ isn't prime.  For example:
$$
x^{2010} - 1 = (x^{67} - 1)(1 + x^{67} + x^{2\cdot 67} + \cdots x^{29 \cdot 67})
$$
This nicely confirms some of your results so far.