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Can you find digits that make the equations true in the following alphametic puzzle (cryptarithm)?

$$RE + MI = FA$$ $$DO + SI = MI$$ $$LA + SI = SOL$$

  • Zero may be a possible variable
  • No one digit may be represented by more than one variable
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What have you tried so far? –  KReiser Aug 23 '12 at 3:37
Can we assume no leading 0's and no digit being represented by more than one letter? –  mathguy Aug 23 '12 at 3:42
Safe to assume leading 0's, but no one digit can be represented by more than one letter. –  user38506 Aug 23 '12 at 4:20
The usual rules for these are no leading zeros and all letters represent distinct digits. –  Ross Millikan Aug 23 '12 at 4:32
Well even if it was a rule that it could be used, I don't think it is possible to have 0 as one of them anyways. –  user38506 Aug 25 '12 at 21:55

1 Answer 1

From the second equation you have $O=0$. Also, since $SOL$ is the sum of two digit numbers, it follows that $S=1$.

Thus, so far we have

$$RE+MI=FA$$ $$D0+1I=MI$$ $$LA+1I=10L$$

From the last equation we get $L=8$ or $L=9$. Also, we know $D+1=M$ from the second.

Now, if each letter is a different digit, combining $E+I=A$ or $E+I=1A$ from the first equation with $A+I=L$ and $L \in \{ 8, 9 \}$, and studying the four possible cases should lead to the solution.

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How do you know O = 0? –  user38506 Aug 23 '12 at 4:18
Look at the last digit of $DO+SI=MI$. –  N. S. Aug 23 '12 at 4:31
@AaronWalker: We know $O=0$ from the ones digit of the second equation, which says $O+I=I$ (possibly, though not in this case, with a carry). This is one of the first things to look at when solving cryptarithms. –  Ross Millikan Aug 23 '12 at 4:31

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