Beginner puzzle

This puzzle is intended to be suitable for people who are new to puzzle solving.

Clarification: Both experienced solvers and new solvers are welcome to post solutions to this puzzle.

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.


This puzzle comes from a UK Junior Mathematical Olympiad.

  • 7
    $\begingroup$ I am a noob at puzzles so this puzzle is perfect for me! $\endgroup$
    – PDT
    Commented Apr 6 at 9:58

2 Answers 2


I believe there are only

2 solutions:

J=1, I=4 or J=2, I=7, M=5, O=0


O can potentially be 5 or 0 because 3xO leads to O as the only or final digit. But it can’t be 5 because that would make no available digits for M since it would be 3M+1 = M as the only or last digit.

Therefore O is 0 and therefore M cannot also be 0 since O is 0 and so the only option left is then 5.

So we have J50x3 making (3J+1)50. Only 1 and 2 can be J since the other numbers will exceed two digits if multiplied by 3. This then leaves 2 possibilities for I: either 4 or 7.


Different explanation but same solution as PDT.

We can do it algebra style
$300 \times J + 30 \times M + 3 \times O = 100 \times I + 10 \times M + O$
Reducing what can be reduced on both sides we get
$300 \times J + 20 \times M + 2 \times O = 100 \times I$
this means that $300 \times J + 20 \times M + 2 \times O$ is divisible by 100.
But since $300 \times J$ is divisible by 100 it means $20 \times M + 2 \times O$ is also divisible by 100.
this can be possible if both M an O are zero, but this is not allowed or if $M = 5$ and $O = 0$
having these 2 digits, and putting them back int the original equation we get that
$300 \times J + 100 = 100 \times I$
or $3 \times J + 1 = I$.
SO J can be 1 which means $I = 4$ or
$J = 2$ and $I = 7$


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