Solving Word Arithmetics

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Solving Word Arithmetics

The name "word arithmetic" is used in Dell puzzle magazines and other books and magazines. It is a form of "cryptarithm." A word arithmetic is a long division problem, with letters in place of the actual numbers. Each letter stands for a different digit. When you are done, the letters, when listed in numeric order, spell out a coherent word or phrase. In this article, I will ignore that feature (even though it may provide clues toward the solution), and just concentrate on the math puzzle (which normally has a unique solution). Other cryptarithms may involve operations other than division, or some of the digits may not be represented by a letter but by a blank or question mark to increase the difficulty.

Here is a typical Word Arithmetic problem (except that you don't see any familiar words in it):

         ABC
       -----
    DE)AFCGH
        DE
       ---
        IDG
        ICD
        ---
         FJH
         FGA
         ---
          JB

The solution depends on a few observation about long division, multiplication, subtraction, and the base-10 number system. Here are a few such observations (using the notation of Word Arithmetic):

Properties of multiplication:
1.  ABC x 0 = 0
2.  If ABC x D = ABC then D = 1
3.  If ABC x D = EF0 then (since you have 
already substituted zero for another letter)
    a. C = 5 and D is even or 
    b. C is even and D = 5
4.  If ABC x D = EFC then (assuming you've 
eliminated D = 1 and C = 0) either
    a. C = 5 and D is odd or 
    b. D = 6 and C is even
5.  If ABC x D = EF1 then (since 1 is already 
acounted for) either
    a. C = 3 and D = 7 or
    b. C = 7 and D = 3
6.  If ABC x D = EFG and G is odd, then C 
and D are both odd
7.  If ABC x D = EFG and G is even, then C 
or D is even (possibly both)
8.  If ABC x D = EFG and D is odd (but not 
equal to 5), then when you determine G and 
D, you can uniquely determine C
9.  If ABC x D = EFG then E > A
10.  If ABC x D = EFGH then E < A
11.  If ABC x D = EFG then A < 5
Properties of subtraction (remembering that 
there may be "borrows" or not):
1. a. If ABC - DEF = GHC then F = 0
    b. If ABC - DEC = FGH then H = 0
2.  If ABC - DE = FG then A = 1
3.  If ABC - DEF = GHF then C is even (2F = C or 10+C)
4.  If ABC - DEF = GH then A = D + 1
5.  If ABC - DEF = GHI then A > D and A > G
6. a.  If ABC - DBE = FGH then G = 0 or 9
    b.  If ABC - DEF = GBH then E = 0 or 9
7.  If ABC - DE = F then A = 1, B = 0, D = 9
8.  AxB=?D and AxC=?D. Then A=5 or even. It should be obvious when A=5 (D=0), so
the rest of the time A is even and D is even. Also, one of B and C is even and
the other is odd.

Also, see my comments on even and odd, at Word Arithmetics - Part 6.

Let's try to solve my problem, from the top of the page:

         ABC
       -----
    DE)AFCGH
        DE
       ---
        IDG
        ICD
        ---
         FJH
         FGA
         ---
          JB

OK, first we notice that A=1, since A x DE=DE. We could have also noticed that AFC - DE=ID, which also tells us that A=1. Next, we notice that G=0, as FJH - FGA=JB. This would tell us that G=0 or 9, but J > G, so G=0. Next, we see that either C=3 and E=7 or C=7 and E=3 (C x DE=F01). It helps if we list these two possibilities:

    C = 3 or 7
    E = 7 or 3

This leads to:

    D = 6 or 4
    J = 4 or 6

because 1FC - DE=ID and ID0 - ICD=FJ. Now we see that D > C (IDG - ICD=FJ), so we can eliminate the right-hand possibility in each of the above 4 equations (C=3 or 7 ... J=4 or 6). F becomes 2 (I60 - I36=F4). And, I becomes 5 (123 - 67=I6). Substituting all of these digits into the diagram:

We can solve for B (B=536 / 67). Then we easily deduce H, the last unknown letter.

The four equations that we listed above (C=3 or 7 ... J=4 or 6) represent the real secret to solving this kind of puzzle. Let's say that, in some puzzle, ABC x D=EFC. Then we see from my list of observations that either:

    a. C = 5 and D is odd or 
    b. D = 6 and C is even

We can list this as:

    C = 5, 5, 5, 2, 4, 8
    D = 3, 7, 9, 6, 6, 6

This gives us 6 columns of numbers, which we can extend downward, by figuring out the consequences of the numbers in each of these columns. In a particularly difficult puzzle, we may end up with many more columns. As an example, lets say we have the above 6 columns. But then we deduce that H is either 0 or 9. We can extend the 6 columns to 11 columns:

    C = 5, 5, 5, 2, 4, 8, 5, 5, 2, 4, 8
    D = 3, 7, 9, 6, 6, 6, 3, 7, 6, 6, 6
    H = 0, 0, 0, 0, 0, 0, 9, 9, 9, 9, 9

You are exploring all of the possible permutations.

See my other articles on Word Arithmetics:

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