Why does the Divisibility Rule of 11 work?

I am sure you all know the Divisibility Rule of 11. Let me put it down here, just in case you forgot:

If you have a 10-digit number like abcdefghij (or 8346834623, for instance), then you add the alternate digits. So, you get (a+c+e+g+i) and (b+d+f+h+j) as two different sums. In our example number, these sums would be 8+4+8+4+2=26, and 3+6+3+6+3=21. Next, you find out the difference of the two sums, that is, 26-21=5.

If this difference is divisible by 11, or is zero, the complete number is divisible by 11 too.

Now, this was only a refresher. But the BIG question is: why does this rule work?

Lets expand the number abcdef (lets take a 6-digit number this time) as:

a x 10^5 + b x 10^4 + c x 10^3 + d x 10^2 + e x 10 + f x 1

When you divide each of the multipliers by 11, you would get 10, 1, 10, 1, 10 and 1 alternately. Now, we know that when you divide "A" and "B" separately by 11, to get the two remainders as "a" and "b" respectively, then you would get a remainder of "axb" when you divide "AxB" by 11. The same hold true for "A+B" too!

Also, a remainder of 10 can be written as -1 (Divisor-Remainder). So the remainder sequence can be re-written as -1,+1,-1,+1,-1 and +1.

So, the remainder of "abcdef" divided by 11 (lets call it abcdefR11) can also be written as:

aR11 x (-1) + bR11 x (+1) + cR11 x (-1) + dR11 x (+1) + eR11 x (-1) + fR11 x (+1)
= (b+d+f)R11 +(-1)x(a+c+e)R11
= [(b+d+f)-(a+c+e)] R11

This is how the Divisibility Rule of 11 is "derived"! You should be able to "explain" the Divisibility Rules of 3 and 9 in a similar manner now! Lets see how many of you can explain them, and earn some coffee with me :-)

Cheers!