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Word Arithmetics - Part 6

© Copyright 1998, Jim Loy

See my other articles on Word Arithmetics:

• Solving Word Arithmetics
• Advanced Word Arithmetics
• Word Arithmetics - part 3
• Word Arithmetics - part 4
• Word Arithmetics - Advice By EMail
• Word Arithmetics - Part 6
• Word Arithmetics - Part 7
• Word Arithmetics - Part 8 (a tough one)

In my articles on Word Arithmetics, I have mainly showed a general method for the very toughest problems. This method involves showing the possible permutations.

But, with one or two insights, you can often avoid that cumbersome method, and just start solving the puzzle. Or, at least you can keep it to a small list of permutations.

In the previous article, Bill Smith showed us two such insights. Here, I will use his second idea, identifying consecutive numbers.

             ABC
         -------
    DEFG)HDICGBF
         HIEHI
         -----
           AGDB
           DEFG
           ----
           BBACF
           BBDEF
           -----
              DJ

This is a puzzle that (as a self-imposed task) I solved by just filling in the solution. In other words, I didn't show my work, doing that in my head. But here, I show my work:

1. J=0 and B=1.
2. We have three letters in sequence IDA (I is one less than D and D is one less than A).
3. In the middle subtraction, we have D-F=A, with or without a borrow. There has to be a borrow, because 1-G=C to the right. So F=8.
4. In the top subtraction, we have I-E=A, with or without a borrow. With no borrow, E is 8. But F is 8. So, there is a borrow, and E=7.
5. G-7=1, with a borrow. G=9.

The rest is easy.

Clues often tell you if a letter is even or odd.

• C-A=A, on the right side of a subtraction (or elsewhere in a subtraction when there is no borrow), shows that C is even. C-A=A, where there is a borrow, shows that C is odd.
• 5 times an even number ends in 0. 5 times an odd number ends in 5.
• 6 times an even number ends in that same even number.
• If AxB=C (...A x ...B=...C) and DxB=C, then B is even (assuming we have eliminated B=5 already). This results from that fact that if B is odd (and not 5), then C is unique. There are not two different multiplications (AxB=C and DxB=C) that can produce C.
• even-even=even, even-odd=odd, odd-even=odd, and odd-odd=even. Borrows reverse each of these.
• even x anything=even, odd x odd=odd.
• There are five even numbers and five odd numbers. If you've identified five letters as being even, then the rest are odd.

Is that useful information? Yes, often. It helps you eliminate quite a few possibilities. If you know that A is 7 or 8, and then see that A is even, then A is 8. If you see even-even=odd, then you know there was a borrow. There are other uses for this information.

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