The sum of ABCD and DCBA is always divisible by 11, where A, B, C and D are digits of a number. 
I understood that, ABCD = A(1000) + B (100) + C(10) + D(1)
 DCBA = D(1000) + C(100) + B(10) + A (1)
Then ABCD + DBCA = A(1001) + B(110) + C (110) + D(1001)
 =11[91A + 10B +10C +91D] which is divisible by 11.

I want a generalized proof for the above problem. Also, I want why it is applicable only for 11 and why it is valid for even digit of numbers?
Thanks in advance