Prove that any palindrome with an even number of digits is divisible by 11
Hint: Notice that $10^k=(-1)^k \pmod{11}$. So, for $k$ odd and $k'$ even, $10^{k} + 10^{k'}=0\pmod{11}$.
Hint: Notice that $10^k=(-1)^k \pmod{11}$. So, for $k$ odd and $k'$ even, $10^{k} + 10^{k'}=0\pmod{11}$.