Proof is by induction.

Every 2 digit palindrome is obviously divisible by 11.

Suppose every 2n digit palindrome is divisible by 11.

Consider any 2(n+1) digit palindrome P. The central 2n digits define a 2n

digit palindrome PP divisible by 11, so write:

P=PP*10 + (10^(2n+1)+1)Q

where Q is the most (and least) significant digit of P. Then P is divisible by

11 because PP is and so is 10^(2n+1)+1.

Then the proof follows by mathematical induction.

RonL

(You may feel the need to prove that 10^(2n+1)+1 is divisible by 11.

This would proceed along the lines of observing that 10=-1 mod 11

so any odd power of 10 is also congruent to -1 modulo 11, which is

equivalent to any odd power of 10 plus 1 is divisible by 11)