Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am referring to puzzles like these, where every letter represents a unique number (0-9):

    ERGO       MAD       LYNDON
 *   SUM   *   MAN      *     B
 -------   -------      -------

Are there any useful tips for solving these? In similar addition alphametics, you usually have to look for carry-on and you continue from there.

But what should I look for in multiplication alphametics? Should I write them as shown above and use long multiplication to solve this, or write them in a regular a * b = c format?

Right now I have no idea where to start - what to look for that could help me, and what technique I should use to continue solving this.

I am looking for tips because I'm clueless and couldn't find any info about how to solve puzzles like these. Any help is appreciated.

share|improve this question

1 Answer 1

up vote 1 down vote accepted

Canonically, I don't know of any 'short cuts' for problems such as this. In general it's just trial-and-error along with some deductive elimination and a lot of branch-and-bound search. I do recommend keeping them in the long-multiplication format, since it makes the bookkeeping a bit easier.

As an example, consider the last problem. Since N*B=N (mod 10), then either B=1 (which can't be the case because the result is longer than the multiplicand) or N has no multiplicative inverse mod 10. This immediately eliminates most of the possibilities for the pair (N, B), leaving you with either N=0 (in which case the same situation can be repeated with B and O), N=5 and B odd, or N even and B=6. Then the repetition of the digit O upon multiplication by B can be used to narrow down possibilities even further; etc.

Similar analyses can start to whittle down the other problems; in the first you have a similar situation for the last digits, along with the fact (by counting digits) that S*E must be less than 10, which narrows down the possibilties for both sharply. In the second, M can't be 1 or 2 for similar product-length reasons, and you can eliminate M=3 without too much effort: M=3 forces A=1 (by product-length reasons), and the only product 31X*31Y that can produce a last digit of 3 and have enough length is 317*319, whose product 101123 doesn't match the ASYLUM digit pattern.

Above and beyond this sort of thing, though, you're mostly stuck with trial-and-error; partial-product versions (where all of the long-multiplication terms 'below the line' that sum to the final product are shown) tend to be substantially easier because there's more opportunity for this sort of deduction.

share|improve this answer
Thank you for your help! –  tempy Oct 16 '12 at 17:21

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.