Word Arithmetics - part 3

See my other articles on Word Arithmetics:

Here are a couple of puzzles.

```          ABC
-----
DEF)GDACE
EGB
---
ABC
HHE
---
FIE
FAJ
---
CE
```

This one is easy, due to a couple of insights. Start with J=0, and C=5 (F can't be 5). If C=5 then D=1 (5 x DEF is only 3 digits). That's our first insight, D=1. Let's put all that into the diagram:

```          AB5
-----
1EF)GDA5E
EGB
---
AB5
HHE
---
FIE
FA0
---
5E
```

We can now find A. A is even (B+B=A or 1A in the top subtraction), but is not zero. A<5 (5+A=I with no carry, in the bottom subtraction). If A is 2, then H is 1, but D is 1. So A=4, by elimination. The puzzle is simple after that:

```          425
-----
168)71456
672
---
425
336
---
896
840
---
56
```

This one is more difficult.

```          ABC
-----
DEF)EGDHF
DEF
---
AIHH
AGJJ
---
FGF
```

It begins easily, with A=1 and C=0. Notice that H-J=G with no borrow (by any subtraction to the right), and H-J=F where there must have been a borrow. So, F, G, and I are consecutive numbers, and J>H. G-E=I, with or without a borrow, so E=9 or 8. E-D=1, with or without a borrow, so D=8, 7, or 6. That restricts F to the range 2 to 5, and it can't be a 5, so F=2, 3, or 4. We can list all of these permutations, and deduce H and J (with conflicts marked as x):

```   E = 99999988888
D = 88877777766
F = 23423423423
G = 34534534534
I = 45645645645
H = 6xx5x35x3xx
J = x  x 8x x```

That leaves us only one column that works. And we can solve the puzzle:

```          120
-----
794)95734
794
---
1633
1588
---
454
```

There are other ways to solve this. For example, we could have listed all possible permutations of H, J, and G. Most of those can be easily eliminated.