Problem 160

For $\textbf{(A)}$ modulo $3,$
$$\begin{align*} 2^{606} - 1 & \equiv (-1)^{606} - 1 \\ & \equiv 1 - 1 \\ & \equiv 0 . \end{align*}$$
Thus, $2^{606} - 1$ is divisible by $3.$
For $\textbf{(B)}$ modulo $5,$
$$\begin{align*} 2^{606} + 1 & \equiv 2^{{\rm Rem} ( 606, \phi(5) )} + 1 \\ & \equiv 2^{{\rm Rem} ( 606, 4 )} + 1 \\ & \equiv 2^2 + 1 \\ & \equiv 0 . \end{align*}$$
Thus, $2^{606} + 1$ is divisible by $5.$
For $\textbf{(D)}$ modulo $3,$
$$\begin{align*} 2^{607} + 1 & \equiv (-1)^{607} + 1 \\ & \equiv - 1 + 1 \\ & \equiv 0 . \end{align*}$$
Thus, $2^{607} + 1$ is divisible by $3.$
For $\textbf{(E)}$ modulo $5,$
$$\begin{align*} 2^{607} + 3^{607} & \equiv 2^{607} + (-2)^{607} \\ & \equiv 2^{607} - 2^{607} \\ & \equiv 0 . \end{align*}$$
Thus, $2^{607} + 3^{607}$ is divisible by $5.$
Therefore, the answer is $\boxed{\textbf{(C) }2^{607} - 1}.$
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
~MrThinker (LaTeX Error)