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Using the following identity: $11 \vert (10^m + (-1)^{m+1}), m \in \mathbb{N}$

Prove the divisibility rule for $11$. (If $11$ divides the alternating sum of digits in a number $n$, then it divides $n$.)

I find this easy to do using modular arithmetic but am struggling in using this identity to prove it.

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  • $\begingroup$ That's what I used to prove it earlier but it doesn't make use of the identity, or at least I am missing how it does $\endgroup$ – Uq'''12wn1F12u2x3uW31H1JBk9m Mar 21 '17 at 21:10
  • $\begingroup$ Is this the idendity ? : $$11|10^m+(-1)^{m+1}$$ $\endgroup$ – Peter Mar 21 '17 at 21:12
  • $\begingroup$ yes that is, I think I will have to take the alternating series of digits and transform it somehow into the sum of digits multiplied with powers of 10 to get n $\endgroup$ – Uq'''12wn1F12u2x3uW31H1JBk9m Mar 21 '17 at 22:19
  • $\begingroup$ Hmmph, I'm not sure why I'd believe the identity without proving it. But .. well, use induction. $\endgroup$ – fleablood Mar 21 '17 at 23:06
  • $\begingroup$ @fleablood I proved the identity earlier using induction. Now I have to prove the divisibility rule for 11 using that identity! $\endgroup$ – Uq'''12wn1F12u2x3uW31H1JBk9m Mar 21 '17 at 23:17
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So "If 11 divides the alternating sum of digits in a number n" means if

$n = \sum_{i=0}^m a_i10^i$ then this is saying:

$11| \sum_{i=0}^m a_i(-1)^i$

So that means $11|\sum_{i=0}^m a_i(-1)^i - a_m(10^m + (-1)^m)= \sum_{i=0}^{m-1}a_i(-1)^i - a_m10^m$.

So that means $11|\sum_{i=0}^{m-1}a_i(-1)^i - a_m10^m - a_{m-1}(10^{m-1} + (-1)^{m-1}) = \sum_{i=0}^{m-2}(-1)^i a_i - \sum_{i= m-1}^m a_i10^i$

And via induction:

$11|\sum_{i= 0}^{m-k-1}(-1)^ia_i - \sum_{i=m-k}^m a_i*10^i$.

Keep going to get:

$11|- \sum_{i=0}^m a_i10^i= -n$ and so $11|n$.

Of, course, I don't know why on earth I'd believe the identity without proving it first.

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