Hive account@kenchung I was going to use a similar method. However, you can actually cut down what M I is even more. Im all for cutting down as much work as you can.

given you got 1300MI3, you can also say that MI can not add up to 2,5,8,11,14, or 17 because the other numbers add up to 5. And we know that any number who's sum of digits is divisible by 3 is also divisible by 3. Combine that with no repeating numbers, the set to try for MI becomes much smaller.

The set of possible solutions to check becomes {24,25,27,28,42,45,46,48,49,52,54,57,58,64,67,69,72,73,75,76,78,79,82,84,85,87,94,96,97}, which is 29% of the work than going through all 100. In fact, you come up with the solution on the second try, thus could have only done 4% of the work needed :)