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© Copyright 1998, Jim Loy
See my other articles on Word Arithmetics:
I received a nice email from Bill Smith, giving me some advice for solving Word Arithmetics. His main piece of advice is this:
The letter representing zero can often be found by identifying nine letters that cannot be zero.
He adds that this technique seems to "uniquely identify zero about half of the times when obvious clues are sparse." He gives an example (from my article, Advanced Word Arithmetics), simpifying that solution:
AB
------
CDEF)EAGHBI
EFGFG
-----
DCEJI
DAACD
-----
HGJG
Bill finds the zero in these few steps:
That's all nine non-zeros. I is zero (A grammar checker may object to that sentence). Bill's technique works beautifully here. And that actually shortens the solution, quite a bit. In other puzzles, some more subtle reasoning (involving clues like N<M, so M can't be zero) may be necessary to find all nine non-zeros. At the very least, identifying two or three candidates for zero may be a valuable clue. I take my hat off to Bill Smith (if I had a hat, and if I were wearing it).
A second technique of Bill's is to identify pairs of consecutive letters. He points out that (in the same puzzle) EA and DB are consecutive (E is one less than A). This comes from a property of multipiers that begin with 9: A x 9DEF=EFGFG, E will be one less than A. In other puzzles, there are often simpler ways of identifying consecutive digits, like M-N=zero in which M is one greater than N, and there was a borrow. Sometimes you see M-N=1, and maybe you have already deduced that there was no borrow. Anyway, Bill points out that the consecutive letters EA appear in the subtraction D9EJ0 - DAA9D=HGJG (the middle digits of the two 5-digit numbers). Depending on whether there was a borrow or not, that makes G 8 or 9. Of course, C is 9, so G is 8. This gives us D=2, and B=3 (they were consecutive). Consecutive letters can give you a variety of clues in a variety of situations. Bill spells out these rules (assuming AB are consecutive):
1a. If xAx - xBx = xCx then C = 8 or 9. 1b. If xAx - xCx = xBx then C = 8 or 9. 2a. If xBx - xAx = xCx then C = 0 or 1. 2b. If xBx - xCx = xAx then C = 0 or 1.
I would like to add that such ideas do not just apply to consecutive digits. Let's say that we know that A-B=5 (or 10+A-B=5), on the right end of some subtraction, and also that xBx-xCx=xAx, then we can deduce that C=5 or 4.
Bill points out that in Dell Math Logic Problems Number 7, Word Arithmetic #42 had two solutions (only one of which spells out a readable phrase). I think I saw something like that, when I was a child, maybe not in a Dell magazine. It has happened every once in a while in Cross Sums and other puzzles. I remember one Cross Sum with about 20 solutions. Cross Sum #4, in the May 1998 Math Puzzles And Logic Problems, has four solutions. I found one puzzle (not Dell) that an author implied had only one solution, which had over 10,000 solutions, that I had my computer print out.