Divisibility Trick for 11 I am trying to prove the 11 alternating sum divisibility trick. I know that $10\equiv -1\pmod{11}$ so for every power of $10$ in a number, we should be able to substitute in $(-1)$ like so:
$$ a(10^n) + b(10^{(n-1)}+\cdots+ c(10) + d \equiv a(-1)^{n} + b(-1)^{n-1}+\cdots+c(-1)+d\pmod{11}$$
What I am having trouble understanding is the fact that starting from the left, we are taking an alternating sum of the digits $(a - b + \cdots$ ). Since each power of $10$ will always become a power of $(-1)$, how can we always use an alternating sum from the left side? That seems like it would change the underlying nature of what is happening with the mod depending on whether we have an even or odd number of digits in the number. For example, when we have a three digit number $a(-1)^3+b(-1)^2+c = -a+b-c$ but according to the rule we have $a-b+c$ verses when we have a four digit number $a(-1)^4+b(-1)^3+c(-1)^2+d = a-b+c-d$ which works with the rule. 
Am I missing something obvious? Do I possibly just have the alternating sum rule written incorrectly in my notes? 
Thanks!

Why are you shifting the variable names.  You can but its confusing.  It=n $a(-1)^2 + b(-1)^1+c=a-b+c$ you have $a,c$ in the odd positions and $b$ in the even position.
But with $a(-1)^3+b(-1)^2+c(-1)^1 + d= -a+b-c+d$ you have $a,c$ switching to the even position, while $b$ switches to the odd.  And $d$ is new in the odd.
For less confusing to list from the right.
$a = a$ ($a$ in odd, nothing in even [divisible by $11$ if and only if $a = 0$])
$b(-1)^1 + a = -b + a$  ($a$ in odd; $a$ will always be in odd.  $b$ in even.)
$c(-1)^2 + b(-1)^1 + a = c - b + a$ ($a,c$ in odd; $b$ in even.)
$d(-1)^3 +c(-1)^2 +b(-1)^1 + a = -d + c -b+a$ ($a,c$ in odd; $b$ in even.... and so on.)
