smc

For $\text{(A)}$ modulo $3,$ $$\begin{array}{rl}{2}^{606}-1& \equiv (-1{)}^{606}-1\\ & \equiv 1-1\\ & \equiv 0.\end{array}$$ Thus, $2^{606}-1$ is divisible by $3.$ For $\text{(B)}$ modulo $5,$ $$\begin{array}{rl}{2}^{606}+1& \equiv 2^{\mathrm{Rem}(606,\varphi (5))}+1\\ & \equiv 2^{\mathrm{Rem}(606,4)}+1\\ & \equiv 2^2+1\\ & \equiv 0.\end{array}$$ Thus, $2^{606}+1$ is divisible by $5.$ For $\text{(D)}$ modulo $3,$ $$\begin{array}{rl}{2}^{607}+1& \equiv (-1{)}^{607}+1\\ & \equiv -1+1\\ & \equiv 0.\end{array}$$ Thus, $2^{607}+1$ is divisible by $3.$ For $\text{(E)}$ modulo $5,$ $$\begin{array}{rl}{2}^{607}+3^{607}& \equiv 2^{607}+(-2{)}^{607}\\ & \equiv 2^{607}-2^{607}\\ & \equiv 0.\end{array}$$ Thus, $2^{607}+3^{607}$ is divisible by $5.$ Therefore, the answer is $\boxed{2^{607}-1}.$