I see Identify unique Verbal Arithmetic Puzzles deals with trying to deal with uniqueness when creating Verbal Arithmetic puzzles, but I have a question that deals with something earlier in the creation process, namely how does one go about picking words for the puzzle to begin with?

  • I could start with any phrase with word lengths roughly the right length and hope that it works, but that is obviously not very efficient.
    • So as an example, I know that LYNDON x B = JOHNSON is one, and I could try other U.S. presidents (like WARREN x G = HARDING), but it's basically like sifting through dirt trying to find gold--there's a low chance of success.
  • Another approach would be to start with the arithmetic equation (and then find words to fit the equations), but the problem there is that the search space there is extremely big. There might be patterns to look for, but I'm not aware of any that would narrow down the possibilities to something that isn't still overwhelming.

So, is there some strategy that makes finding words for these puzzles easier, or am I stuck with the naive approaches?

  • $\begingroup$ This seems to be the same question as mine... The other question's answer though really one said "pick words and have your solver tell them if they are right" so there isn't really an answer there yet. $\endgroup$
    – kaine
    Commented Jun 20, 2014 at 14:44
  • $\begingroup$ @kaine I want to say not exactly, not the least because I'm not that worried about uniqueness (as long as there isn't some outlandishly high number of solutions, I'm fine). Though I admit, there could potentially be one strategy that answers both of our questions. $\endgroup$ Commented Jun 20, 2014 at 15:39

1 Answer 1


If there is no worry about multiple solutions (which I feel is required for it to be a verbal arithmetic problem), find words is simple as long as one follows the following rules:

  • There are no more than 10 different letters. 7-9 is probably the most useful in order to limit too many answers.
  • Count the appropriate number of digits. If you want $5$ letters times $1$ letter, the result can only have $5$ or $6$ letters as $99999*9=899991$ and $10000*1=10000$ which are the extreme examples. Use those extremes to determine whether the equation make sense.
  • Shorter words (3-5 letters) work better than longer ones. You are setting up a system of simultaneous equations. The fewer equations, the more likely they will work out but the more likely there will be duplicate answers.
  • Have a computer based solver ready to rapidly check whether a solution exists. If follow the above rules, most sets of words will yield at least one solution.
  • $\begingroup$ What about verbal arithmetic problems using addition? I only happened to use multiplication ones as examples (I'm not limiting myself to just those) $\endgroup$ Commented Jun 20, 2014 at 16:01
  • $\begingroup$ It all still applies exactly the same way. $9999+9999=18999$ means $4$ letters plus $4$ letters can have an answer with at most $5$ letters but must have at least $4$ letters. $\endgroup$
    – kaine
    Commented Jun 20, 2014 at 16:06
  • $\begingroup$ I guess I was wondering since in the addition case, we're dealing with a larger search space; if the computer based solver still has to be run billions of times over a day then it's not appreciably better than the naive approach $\endgroup$ Commented Jun 20, 2014 at 16:09
  • 1
    $\begingroup$ For any equation with 10 different letters there are only $10!$ number/letter combinations which is less than 3.7 million. For 8 (see rule 1) there are only $8! \approx 40,000$. For a computer that is very quick and very few. Addition vs. multiplication does not change this. $\endgroup$
    – kaine
    Commented Jun 20, 2014 at 16:13
  • $\begingroup$ Actually it is roughly 1800,000 for the 8 case but the argument still applies. $\endgroup$
    – kaine
    Commented Jun 20, 2014 at 16:23

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