Word Arithmetics

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Word Arithmetics - Part 7

See my other articles on Word Arithmetics:

Sometimes you can study a seemingly difficult word arithmetic puzzle for a while, and then get a perfect insight to solve it in one swoop.

Part I - This is a puzzle like that:

            ABC
         ------
     DAE)FEGHBH
         FBDB
         ----
          IADB
          IGFA
          ----
           CJEH
           BCGB
           ----
            FHD

I see one clue right away: AxE=...B, and BxE=...A. I know from doing these puzzles that there are several combinations that work that way, especially when E is 9. But, I see a better clue. AxE=...B and A+E=...B. Off hand, I couldn't think of any combinations that work that way. Let's draw a multiplication table (note that A is not equal to E, also A and E are not equal to zero):

   A =  1 2 3 4 5 6 7 8 9
        -----------------
E = 1 |   2 3 4 5 6 7 8 9
    2 | 2   6 8 0 2 4 6 8
    3 | 3 6   2 5 8 1 4 7
    4 | 4 8 2   0 4 8 2 6
    5 | 5 0 5 0   0 5 0 5
    6 | 6 2 8 4 0   2 8 4
    7 | 7 4 1 8 5 2   6 3
    8 | 8 6 4 2 0 8 6   2
    9 | 9 8 7 6 5 4 3 2

We see that there are only two possible solutions 4x8=8x4=...2 and 4+8=8+4=...2. Now we can use our other clue, BxE=...A to eliminate one of these two possible solutions. And the puzzle is easy from there.

The solution:

Part II - Here is another puzzle, which seems difficult at first:

            AB
         -----
    CDEF)BGHGF
         GDIC
         ----
         EGCJF
         EGEEH
          ----
           EAJ

Any clues? How about these:

AxCDEF is a 4-digit number
A is not 1 or 0
E is not 0
E+E=C (or E+E=C+1)

These facts simplify the puzzle tremendously. They limit A (2, 3, or 4), E (1, 2, 3, or 4), and C (2, 3, or 4):

    C=2234
    E=1112
    A=3422

We can eliminate the last column, as E and A cannot both be 2. So E=1. AxF=?C, which eliminates the third column (where A is even and C is odd), so C=2. And we can limit possible values of F:

    C=222
    E=111
    A=344
    F=438

A+E=J or A+E=J+1, giving us possible values of J, and those give us G and H:

We can eliminate the last column. And if H is odd then F cannot be even. That eliminates two more columns:

    C=22
    E=11
    A=44
    F=33
    J=56
    G=78
    H=87

We now know C, E, A, and F. Let's put those into the puzzle:

            4B
         -----
    2D13)BGHG3
         GDI2
         ----
         1G2J3
         1G11H
          ----
           14J

I is 5, which makes J=6, G=8, and H=7:

Finally, B=9 and D=0. And that solves the puzzle.

Part III - This one is easy. Try it. You may have trouble at first. But there are many clues:

            AB
         -----
    CDDE)FGABD
         HGIE
         ----
          JIBD
          JFEE
          ----
           ABD

The following clues should be obvious: E=0, J=9, B=5, A is even, and D is even. Also C=1, as 5xCDDE is a four digit number. D=8, as no other number works for 5x1DD0=9F00. A<5 as H<J=9. So A=2 or 4. A=4 doesn't work with Ax1880=HGI0. So A=2. And the rest is simple.

Part IV - This one is very easy, if you notice a certain clue. Try it:

             ABC
         -------
    DABE)EDBFFGC
         EHGCE
         ------
          HABIG
          HBCAG
          ------
           HFHJC
           HGEHJ
           -----
             GFC

First of all, J=0, C=5, A=6, and E is even. The puzzle looks a little daunting after that. But, E can be deduced immediately. 6xBE=...5E. So 6xE must have an odd carry. 6x2 is the only such number left. So E=2. Then B=4. The rest of the puzzle is easy.

Comment: A few months ago, Dell provided a page of Word Arithmetics that did not spell words or sentences, like mine, as an experiment for their serious Word Arithmetic fans. Unfortunately, most of the puzzles were ridiculously easy.

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