B1. In this word-sum, each letter stands for one of the digits $0-9$, and stands for the same digit each time it appears. Different letters stand for different digits. No number starts with 0 .
Find all the possible solutions of the word-sum shown here.
SolUTION
To find $O$, we add three $O$ s to find a number that has an $O$ in the units place. Hence we require $3 O=O$ or $3 O=O+10$ or $3 O=O+20$. Therefore, $2 O=0$ or $2 O=10$ or $2 O=20$ respectively. The only digits which satisfy any of these conditions are 0 and 5 .
If $O=0$, there are no tens to "carry" to the middle column and so we are seeking a digit for $M$, for which three $M$ s have an $M$ in the units place. Therefore, using a similar argument as the one for $O, M$ is 0 or 5 . However, $O=0$ and so $M=5$ (as different letters stand for different digits).
If $O=5$, the total of three $O$ s is 15 and so we have a ' 1 ' to carry to the middle column. Hence, we require $3 M+1=M$ or $3 M+1=M+10$ or $3 M+1=M+20$, which implies $2 M+1=0$, $2 M+1=10$ or $2 M+1=20$ respectively. However, there are no digit values of $M$ which satisfy these conditions.
Hence, $O=0$ and $M=5$. If $J=1, I=4$. If $J=2, I=7$. If $J>2,3 \times$ ' $J M O$ ' does not give a 3-digit number and so there are only 2 possible values for $J$.
The only solutions are ' $J M O$ ' $=150$ with ' $I M O$ ' $=450$ and ' $J M O$ ' $=250$ with ' $I M O$ ' $=750$.