SAUDI ARABIAN MATHEMATICAL COMPETITIONS

a. We will show that for all $k>0$, the number $n=(pk+1)(p-1)$ satisfies $p\mid n\cdot 2^n+1$. Indeed, by Fermat's Little Theorem, we have $$2^{p-1}\equiv 1(\mathrm{mod}p)\text{ so }{2}^n=2^{(pk+1)(p-1)}\equiv 1(\mathrm{mod}p).$$ And then $$n\cdot 2^n+1\equiv (pk+1)(p-1)\cdot 1+1=p^{2}k-pk+p\equiv 0(\mathrm{mod}p).$$ Since there are infinitely many numbers of the form $(pk+1)(p-1)$, we get the conclusion.

b. Notice that $n$ is periodic modulo $3$, with a period of $3$, and $2^n$ is periodic modulo $3$, with period $2$. Hence, $n\cdot 2^n+1$ is periodic, with period (at most) $6$, and only the first $6$ positive integers need to be analyzed. The answer is $n=6k+1$ or $n=6k+2$.