Divisibility Rule for 11 Using Identity

Question

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.

Answer 1

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.