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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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    Sorry, I'm not sure, can you elaborate?2012-03-30
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    @DominickGerard: What are you not sure about?2012-03-30
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    i'm not sure that I see the pattern, kind of confused about how to find $b_n$2012-03-30
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    @DominickGerard: $b_n$ is the n digit number where all digits are ones. The above rule says that if you take the sum of the digits which appear at even positions and take the sum of digits at odd position and subtract, the difference should be be divisible by 11 for the original number to be divisible by 11. For an eg of the application of the rule: 121. You take 1 + 1 -2 = 0. Which is divisble by 11. So 121 is also divisible by 11 If you consider 1234, (1+3) - (2+4) = -2, is not divisible by 11. So 1234 is not divisible by 11 . Now try it with $b_n$.2012-03-30
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    my best guess is that the $b_n$ with an even amount of digits are divisible by 11, but an odd amount are not (ie 1111 is divisble by 11, but 11111 is not). Is there a reason for this? or an eloquent way to put this?2012-03-30
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    How do I prove this is true?2012-03-30
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    @DominickGerard: I just gave you a method (based on the first part of your question) to determine if a number is divisible by 11 or not. Why don't you try using that?2012-03-30
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    I'm sorry, perhaps I do not fully understand the notation?2012-03-30
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    @DominickGerard: Did you read my comment of 43 minutes ago? You did not understand that? Which parts?2012-03-30
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    I understand that comment, just confused as how to turn that into a formal proof? My apologies again, I feel like I should be seeing this.2012-03-30
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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