With regards to a general solution for an n digit number, you can, using my previous logic, state that the digits of the number are a, b, c, d, ..., y, z in order. You can also write the number as an expression, in the form:
10^(n-1)*a + 10^(n-2)*b + 10^(n-3)*c + 10^(n-4)*d + ... + 10y + z
The permutation can also be written in this form as:
10^(n-1)*b + 10^(n-2)*c + 10^(n-3)*d + 10^(n-4)*e + ... + 10z + a
(by just shifting all of the digits to the left)
By adding these two expressions together and then simplifying, we get the expression:
a(10^(n-1) + 1) +b(10^(n-2) + 10^(n-1)) + c(10^(n-3) + 10^(n-2)) + ... + z(10 + 1).
Now, let us take the expression of 10^(n-2) + 10^(n-1). If we express this in a numerical form, we will get the number 11 followed by (n-2) zeros, as 10^(n-2) is just 1 followed by (n-2) zeros and 10^(n-1) is just 1 followed by (n-1) zeros. Thus we can express 10^(n-2) + 10^(n-1) as 11 * 10^(n-2). In fact, we can do this with every part of the expression above, and can simplify yet again to:
a(10^(n-1) + 1) + b(11*10^(n-2)) + c(11*10^(n-3)) ... + z(11*1)
Or, factoring out the 11:
a(10^(n-1) + 1) + 11(10^(n-2)b + 10^(n-3)c + ... + z )
So, as we wish 11 to go into all of the above expression, all we need to do is prove 10^(n-1) + 1 is divisible by 11. If we were to write this number out, it would be in the form 1 followed by (n-2) zeros followed by a 1. The divisibility rule for 11 is to take the alternating sum of the digits in the number, read from left to right. If that is divisible by 11, so is the original number. There are 2 potential cases here: the digits sum to 0 or to 2, as we can either have the expression 1 - 0 + 0 - 0 ... + 0 - 1 = 0 OR 1 - 0 + 0 - 0 ... - 0 + 1 = 2.
As 0 is a multiple of 11 (11*0) but 2 is not, we need to only find cases where 0 occurs. This ONLY happens when n is even, as when n is even, the last alternating sign HAS to be a -. Thus, 10^(n-1) + 1 is divisible by 11 iff (IF AND ONLY IF) n is even!
Thus, as two multiples of 11 sum to a multiple of 11, we can prove that for an n digit number, WHEN N IS EVEN, if you add the original number to it's permutation you will always get a multiple of 11. Finally, QED