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© Copyright 1998, Jim Loy
See my other articles on Word Arithmetics:
Here are a few puzzles which look difficult, but which are relatively easy once you have a certain insight.
1.
ABC
------
DEF)GBEFDH
GADB
----
IDDD
IHJA
----
ICFH
IJHD
----
AEF
The clues are fairly sparse here. There's only one really useful clue, that I notice. AxF=...B and BxF=...A (the first two multiplications). I recall that this happens when F=9. Does it happen for other numbers (where A, B, and F are three distinct numbers)? Let's look at a multiplication table:
A= 0 1 2 3 4 5 6 7 8 9
F= 0 0 0 0 0 0 0 0 0 0 0
1 0 1 2 3 4 5 6 7 8 9
2 0 2 4 6 8 10 12 14 16 18
3 0 3 6 9 12 15 18 21 24 27
4 0 4 8 12 16 20 24 28 32 36
5 0 5 10 15 20 25 30 35 40 45
6 0 6 12 18 24 30 36 42 48 54
7 0 7 14 21 28 35 42 49 56 63
8 0 8 16 24 32 40 48 56 64 72
9 0 9 18 27 36 45 54 63 72 81
OK, we see that 2x4=8 and 8x4=32 work, too. This gives us a small number of possible permutations:
A=28234678
F=44999999
B=82876432
Solving the puzzle from there is fairly easy.
2.
AB
-----
CDE)FCGDH
FDCE
----
IFDH
AJDE
----
AEH
E=0. And BxD ends in D, so either D=5 and B is odd, or D is even and B=6. But, D cannot be 5, as then C would be 5 or 0. So, D is even and B=6. Let's put those into the diagram:
A6
-----
CDE)FCGDH
FDC0
----
IFDH
AJD0
----
A0H
We also have C is even, and C>D. There are only three even numbers left, 2, 4, and 8. So, we have 3 possible permutations of C and D:
C = 4 8 8
D = 2 2 4
And we can solve for J and A. And the puzzle is easy from there.
3.
AB
-----
ACD)EFCEG
EBHC
----
IJG
B=0 (for two reasons), and F=1. AxACD=E0HC is 4 digits, with the leftmost digit not being 1. So, A>3, and A is not equal to 5. If A is 4 then AxCD would have to produce a carry of 4, to make B=0. There is no number CD that produces a carry of 4. So, A>5. If A is 9, then AxCD would have to produce a carry of 9, to make B=0. Likewise, there is no CD that produces a carry of 9. So, A is 6, 7, or 8.
If A is 6, then AxCD produces a carry of 4, so C is 7, or 8 (can't be a 6). If A is 7, then AxCD produces a carry of 1, so C is 2. If A is 8, then AxCD produces a carry of 6, so C is 7. Let's list those permutations:
A = 6 6 7 8
C = 7 8 2 7
We can deduce D (which would make E0HC end in C. We find that C can't be 7. I now list that info:
A = 6 6 7 8
C = 7 8 2 7
D = 3 6
Now we just see if 6x683=4098 or 7x726=5082 are consistent with E0HC. We find that both of those are consistent. But, we can solve for I and J, and we eliminate one of these answers, and go on to solve the puzzle.
4.
AB
----
CDE)FGDH
FIG
---
DJH
B=0. AxCDE=FIG, which is only 3 digits. C>D, so C>1. The permutations which make AxCDE be a 3 digit number, are:
A = 3 4 2 2 2
C = 2 2 3 4 4
D = 1 1 1 1 3
For each of these, we have several values which E (which can't be a 5) can be:
A = 33333 4444 2222 2222 222
C = 22222 2222 3333 4444 444
D = 11111 1111 1111 1111 333
E = 46789 3789 4789 3789 789
Solving for some of the other letters, eliminates all but 4 permutations:
* * * *
A = 33333 4444 2222 2222 222
C = 22222 2222 3333 4444 444
D = 11111 1111 1111 1111 333
E = 46789 3789 4789 3789 789
G = 28147 2826 8468 6468 468
J = 3 74 3 5 3753 5 53 75
I = 25 6 4 4 4 46 24
* * * *
For, each of these four permutations, we multiply out AxCDE, to see if we get a consistent 3-digit answer: 219x3=657, 217x4=868, 318x2=636, 419x2=838. Only the first is consistent. And we go on to solve the puzzle.