Prove that the number 3^105+4^105 is divisible with 91 but not with 11.
What are the remainders when powers of 3 are divided by 11?
Looks like there's a pattern there! Now do the same thing for powers of 4:
Similar pattern, repeating in batches of five.
Try to prove why this pattern occurs, then notice that at the 105th stage the sum of the remainders is not 0.