Clearly divisible by $2$, leaving us to show that $997$ divides it. But $10^{900} \equiv 1000^{300} \equiv (-3)^{300} = (-27)^{100} \mod 997$ and $2^{1000} \equiv 1024^{100} \equiv 27^{100} = (-27)^{100} \mod 997$, thus we are done.