Copyright © University of Cambridge. All rights reserved.
'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:
||$C$ and $O$
||$4$ and $7$
||$3$ and $8$
||$5$ 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.View the previous week's solutionView the current weekly problem