Congruence Summation Notation I am given this problem:
Suppose that a positive integer $n$, written in decimal notation, has digits (from left to right) $a_k, a_{k-1}, \ldots, a_0$. So $n = a_k 10^k + a_{k-1} 10^{k-1} + \cdots + a_1 10^1 + a_0$. Prove that $$n \equiv \sum_{i=0}^k (-1)^i a_i \equiv (-1)^k a_k + \cdots - a_3 + a_2 - a_1 + a_0 \quad \text{(mod $11$)}$$
Now apply this result: let $b_n$ denote the number consisting, in decimal notation, of $n$ $1$'s. That is $$
b_n = \underbrace{11 \cdots 1}_n $$
For which $n$ is $b_n$ divisible by $11$?
I am not sure how to approach this, what is the best way to solve this?

$10 = -1 \mod 11$ and so $10^i = (-1)^i \mod 11$
Thus $$\sum_{i=0}^{k} a_i 10^i = \sum_{i=0}^{k} a_i (-1)^i \mod 11$$
Try computing the same for $b_n$ for some $n$. Do you see a pattern?
