jmc

We first look at the hundreds place. Since $E\ne G$, it must be that $E+1=G$ in order to get $G$ in the hundreds place. Since a $1$ is carried over, we have $G+G=10+M$. Now we look at the units place. Either $M+M=M$ or $M+M=10+M$. In the second case, $2M=10+M\Rightarrow M=10$, which is not a possible digit. So it must be that $2M=M$, which is only possible if $M=0$. Now $2G=10\Rightarrow G=5$ and $E+1=G\Rightarrow E=4$. The numerical value of $E$ is $\boxed{4}$. We can check that $450+50=500$, which matches the digits in the addition problem.