I'm dreadful at this sort of thing. Try this link:

http://mathforum.org/library/drmath/view/58510.html

First, it isn't quite true as stated.  The sums can differ by a multiple of 11, e.g. 979 = 11x89, 9+9 = 7 + 11.

Given a number X, construct another number Y as follows: Subtract the sum of the even-position digits (10s, 1000s, ...) from the sum of  the odd-position digits (1s, 100s, ...)

Now look at what happens to Y if you add 1 to X and do it again.  If there's no carry, Y has gone up by 1. If there's a carry from the 1s to the 10s, but no further, the odds total has gone down by 9 and the evens total up by 1, so Y has gone down by 10.  If the carry extends to the 100s then the odds have gone down by 8 (-9+1) and the evens down by 9, increasing Y by 1, and so on.  Y either goes up 1 or down 10.  Adding 2, the result goes up 2, or down 9, or down 20, numbers which differ by multiples of 11.  Adding 11, Y goes up 11, stays the same, goes down 11, or goes down 22, etc.  So when X is a multiple of 11, so is Y.

this is an example of the remainder theorem in algebra.

this states (and can be proved by actual division) that if

p.x^4 + q.x^3 +r.x^2 + s.x + t (" ^ " is raised to power) no end to terms

is divided by x - a

the remainder is p.a^4 + q.a^3 + r.a^2 + s.a + t

This does not sound very helpfull.  But if x = 10 and a = 1 then the long expression is a number in the decimal notation and x - a is 9.

 Hence "casting out nines".  The remainder when a number is divided by 9 is p.1^4 + q.1^3 ....  which is p + q + r + s + t.  If this number itself is greater than 9, than find the remainder when it is divided by a multiple of nine.

Now, if a = -1, so that 10 - a = 11 then we are dividing by 11 rather than nine. So the value of the remainder by the above rule becomes

p. (-1)^4 + q.(-1)^3 + r.(-1)^2 + s.(-1) + t

which equals p - q + r - s + t

If the expression is divisible by eleven, then the remainder will be zero (or at least itself divisible by 11)