Sequence of integers This one should be quite easy, I tried with mathematical induction but the things started to complicate so I would like to see how someone of you there will prove this: 

Prove that $$a_k=(18k+1)2^{18k+1}+1$$ is composite number for every $k\gt 0$.


Go $\pmod 3$ and see what happens. However your mouse over the gray area below for the complete solution.

$$2 \equiv -1 \pmod3 \implies 2^{18k} \equiv 1 \pmod3 \implies 2^{18k+1} \equiv -1 \pmod3$$Also, $$18k+1 \equiv 1 \pmod 3$$Hence,$$(18k+1)2^{18k+1} \equiv -1 \pmod3 \implies (18k+1)2^{18k+1} + 1 \equiv 0 \pmod3$$

