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## 'Sum Total' printed from http://nrich.maths.org/

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.*