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Since no number starts with a $0$, the value in the hundreds column must carry, so $S=1$.

Then, $C+O$ must have units digit $1$. As $S$ is $1$, one of $C$ and $O$ must be at least $2$, so $C+O=11$. In particular, this means there is a carry into the tens column.

In the tens column, since $M+M+1$ is larger than $M$, there must be a carry. Therefore, $M+M+1 = M+10$, so $M=9$.

Since this carries, $U$ must be odd. There are then three possibilities for $U$: $3$, $5$ and $7$. The values of $J$, $C$ and $O$ generated are then shown in the table:

$U$ $J$ $C$ and $O$
$3$ $6$ $4$ and $7$
$5$ $7$ $3$ and $8$
$7$ $8$ $5$ and $6$

$C$ and $O$ could be these numbers in either order, which gives the following six solutions:

This problem is taken from the UKMT Mathematical Challenges.