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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?

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    Are there other ways to solve this problem? I'm truly stuck.2012-03-30

1 Answers 1

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$10 = -1 \mod 11$ 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?

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    @DominickGerard: I am not sure what else I can do to help... Perhaps you can try writing in terms of the $a_i$ and see what $a_2 + a_4 + \dots - (a_1 + a_3 + \dots)$ for the $b_n$ comes out to be. Consider two cases, $n$ even and $n$ odd.2012-03-30