I think such type of questions are best dealt with by Mathematical Induction.
Prove 5^(2n+1)+11^(2n+1)+17^(2n+1) is divisible by 33.
I wonder if there are other ways that can prove this.
My way of proving this:
Let P(n) be "5^(2n+1)+11^(2n+1)+17^(2n+1) is divisible by 33." for all positive integers n.
When n = 1, 5^3+11^3+17^3 = 33*193
Thus P(1) is true.
Assume P(k) is true for some positive integers k.
i.e. 5^(2k+1)+11^(2k+1)+17^(2k+1) = 33N, where N is an integer.
When n = k+1,
5^[2(k+1)+1] + 11^[2(k+1)+1] + 17^[2(k+1)+1]
=5^(2k+3) + 11^(2k+3) + 17^(2k+3)
=25[5^(2k+1)] + 121[11^(2k+1)] + 289[17^(2k+1)]
=25[5^(2k+1)+11^(2k+1)+17^(2k+1)] + 96[11^(2k+1)] + 264[17^(2k+1)]
=25(33N) + 99[11^(2k+1)] - 3[11^(2k+1)] + 264*17^(2k+1)
=33[25N + 3*11^(2k+1) - 11^(2k) + 8*17^(2k+1)]
Since N and k are integers,
25N + 3*11^(2k+1) - 11^(2k) + 8*17^(2k+1) is an integer.
Therefore, 25N + 3*11^(2k+1) - 11^(2k) + 8*17^(2k+1) is divisible by 33.
Thus P(k+1) is true.
By M.I., P(n) is true for all positive integers n.
Another way
The given expression is divisible by 11 using the fact that 5 = -6 (mod 11) and 17 = 6 (mod 11)
Similarly, we can use congruence mod 3 to show that the expression is divisible by 3
Therefore it is divisible by 33