So I was looking something up in another math book of mine, and I came across this intriguing problem.

A number in decimal is palindromic if the digits read the same forward and backward. Prove that:

a) Every palindromic integer with an even number of digits is divisible by 11.

b) Every integer whose base k representation is palindromic and has even length is divisible by k+1.

The proof for (a) is sort of trivial if you remember the divisibility rule for 11. And you can find slightly more interesting proofs online, or try it on your own!

(b) seems more exciting and I haven't thought too much about how that proof would work yet, but just to see it in action, we'll take a number in hexadecimal (base 16) and show that it's divisible by 17.

So take AC33CA which is 10*16^{5} + 12*16^{4} + 3*16^{3} + 3*16^{2}+12*16 + 10 = 11285450 = 17 * 663850.