See my other articles on Word Arithmetics:

```             AB
------
CDEF)EAGHBI
EFGFG
-----
DCEJI
DAACD
-----
HGJG
```

We begin my noticing the column G-G=C, in the middle of the puzzle. C is either 0, or there was a borrow from this column, when C is 9. In the bottom subtraction, DCEJI > DAACD (one of the rules of our long division process). So, C>A and C is 9. We put that into our puzzle:

```             AB
------
9DEF)EAGHBI
EFGFG
-----
D9EJI
DAA9D
-----
HGJG
```

For similar reasons, we notice that A>F. This is interesting, because AxF=G (perhaps with a digit to the left of G), and A-1-F=D (there is a borrow). I see no other obvious clues. So, let's list all of the permutations of A>F, noticing that F cannot be a 1 or a 0 or a 5, while A cannot be a 5:

```  A=344677778888
F=223323462347```

Notice that I also did not list the permutations involving 6 and an even number, as 6xe=...e (6x2=12, 6x4=24, 6x6=36, 6x8=48). We can now solve for G and D, for each permutation. We add these to our table of permutations:

```    * * ** * * *
A=344677778888
F=223323462347
G=682841826426
D=010243205430```

We can now eliminate several of these permutations because D can't be a zero (we don't allow numbers beginning with zero), G cannot equal D (we have two examples of this where G and D=4), and F cannot equal D (we have one example of that where F=D=3). I've put asterisks over those permutations. I would normally cross out these columns. Here, I will put x's where the numbers were. And, we can solve for I (G+D=I, bottom right):

```    * * ** * * *
A=x4x6xx7x8x8x
F=x2x3xx4x2x4x
G=x8x8xx8x6x2x
D=x1x2xx2x5x3x
I= 9 0  0 1 5```

I cannot be a 9 (C=9). D cannot be a 5 (BxF=D). In the lower right column of the puzzle, we notice that D>I, as there is a borrow from the column just to the left of that column. So, we can eliminate three more permutation (leaving just two permutations):

```    *** ** *****
A=xxx6xx7xxxxx
F=xxx3xx4xxxxx
G=xxx8xx8xxxxx
D=xxx2xx2xxxxx
I= x 0  0 x x```

So, we see that G=8, D=2, and I=0. We can put these into the puzzle:

```             AB
------
92EF)EA8HB0
EF8F8
-----
29EJ0
2AA92
-----
H8J8
```

Now we notice that BxF=2. F is 3 or 4. If F=3 then B=4, and if F=4 then B=either 3 or 8. But G=8, so B can't be 8. In the upper subtraction, B-8=J. Using B=4 or 3, we find that J=6 or 5. We can append these to the bottom of our two final permutations:

```    *** ** *****
A=xxx6xx7xxxxx
F=xxx3xx4xxxxx
G=xxx8xx8xxxxx
D=xxx2xx2xxxxx
I= x 0  0 x x
B=   4  3
J=   6  5```

The left permutation has A and J both equal to 6. So, we eliminate that permutation, leaving the right column. We can substitute this into the puzzle:

```             73
------
92E4)E78H30
E4848
-----
29E50
27792
-----
H858
```

Now we can easily solve for E and H, to solve the puzzle:

```             73
------
9264)678130
64848
-----
29650
27792
-----
1858
```

There are more difficult Word Arithmetic problems. The solution to these almost always involves making an even bigger list of possible permutations. Then, logical deductions allow you to eliminate permutations, just as we did above.

There was a shortcut that I didn't mention, above. It involved a more difficult insight.

```             AB
------
9DEF)EAGHBI
EFGFG
-----
D9EJI
DAA9D
-----
HGJG
```

In this diagram, we can deduce that E is one less than A, and D is one less than B. This comes from the nature of numbers beginning with 9, like 9DEF. E is either equal to A (not possible here) or one less than A. The same reasoning applies to D and B.

We could have also tried to solve for H, earlier. 9-A=H, there may be a borrow or not. So, for every value of A, in the permutation table, we would have two values of H (with or without a borrow). This would have doubled the number of permutations, but it may have been easier. For example, let's add H to the bottom of the first table:

```     no borrow      borrow
A=344677778888 344677778888
F=223323462347 223323462347
H=655322221111 544211110000```

As you can see, some of these can be eliminated, right away (H cannot be 0, and H cannot equal A or F). Other permutations will be eliminated later.

This trick (doubling the size of the permutation table, depending on a borrow or not) sometimes comes in very handy.

Let's try a really tough one:

```         A
---
BC)DEF
GHI
---
JJ
```

Not a lot of clues here:

1. A is not a 1 or a 5
2. C is not a 0 or 1 or a 5
3. D is one greater than G
4. H>E
5. J>E
6. A>D
7. A>G
8. A>J
9. B>D
10. B>G
11. B>J
12. G is not 0
13. If C is even, then A is not a 6
14. If A is even, then C is not a 6

That's quite a lot, actually. All word arithmetics yield similar inequalities. But, we usually have more useful clues, so we ignore these clues until we start eliminating permutations. Anyway, we notice that A and B are greater than D. And A, B, and D are greater than G, which is not 0. So, A and B are greater than 2. D<8 and G<7. Let's try the permutations involving A and C:

```  A=3333334444466677777788888999999
C=2467892378937923468923479234678
I=6281478282682441826364262876432```

If we add B to this table, we can solve for G, H, and D (for each column of the resulting table):

```  A=33333 33333 3333 33333 3333 3333 44444 44444 4444 44444 ...
C=22222 44444 6666 77777 8888 9999 22222 33333 7777 88888 ...
I=66666 22222 8888 11111 4444 7777 88888 22222 8888 22222 ...
B=45789 56789 4579 45689 5679 4568 35679 56789 3569 35679 ...
A=... 4444 6666 66666 6666 77777 77777 7777 77777 7777 7777 ...
C=... 9999 3333 77777 9999 22222 33333 4444 66666 8888 9999 ...
I=... 6666 8888 22222 4444 44444 11111 8888 22222 6666 3333 ...
B=... 3578 4579 34589 3578 35689 45689 3569 34589 3459 4568 ...
A=... 88888 8888 88888 8888 88888 99999 9999 9999 9999 9999 99999
C=... 22222 3333 44444 7777 99999 22222 3333 4444 6666 7777 88888
I=... 66666 4444 22222 6666 22222 88888 7777 6666 4444 3333 22222
B=... 34579 5679 35679 3459 34567 34567 4568 3578 3578 4568 34567
```

That's cumbersome. But, I see no other way. We can now solve for G, H, and D:

```    ***** ***** **** ***** !*** **** ***** ***** **** *****
A=33333 33333 3333 33333 3333 3333 44444 44444 4444 44444 ...
C=22222 44444 6666 77777 8888 9999 22222 33333 7777 88888 ...
I=66666 22222 8888 11111 4444 7777 88888 22222 8888 22222 ...
B=45789 56789 4579 45689 5679 4568 35679 56789 3569 35679 ...
G=11222 11222 1122 11222 1222 1122 12223 22233 1223 12233 ...
H=25147 69252 3628 47069 7039 4706 20486 15937 4268 53719 ...
D=      22           33  23     33     4            2  4  ...
!*** **** ***** **** ***** ***** **** **!** **** **!*
A=... 4444 6666 66666 6666 77777 77777 7777 77777 7777 7777 ...
C=... 9999 3333 77777 9999 22222 33333 4444 66666 8888 9999 ...
I=... 6666 8888 22222 4444 44444 11111 8888 22222 6666 3333 ...
B=... 3578 4579 34589 3578 35689 45689 3569 34589 3459 4568 ...
G=... 1233 2345 22355 2345 23456 33456 2346 23366 2346 3446 ...
H=... 5315 5135 28428 3573 26374 07485 3745 52907 6308 4182 ...
D=... 2344 3      4 6    6  4        7    7   4     5   557 ...
***** **** ***!* **** ***** ***** **** **** **** **** *****
A=... 88888 8888 88888 8888 88888 99999 9999 9999 9999 9999 99999
C=... 22222 3333 44444 7777 99999 22222 3333 4444 6666 7777 88888
I=... 66666 4444 22222 6666 22222 88888 7777 6666 4444 3333 22222
B=... 34579 5679 35679 3459 34567 34567 4568 3578 3578 4568 34567
G=... 23457 4557 24557 2347 33456 23456 3457 3467 3567 4567 34567
H=... 53173 2084 73195 9757 19753 87654 8764 0865 2087 2108 43210
D=...   5 8  6     668 3      5 7  45 7         8
```

Again, I have marked each column which contains a conflict, with an asterisk. There are only a few without conflicts. I will list those, now:

```  A=34778
C=89694
I=46232
B=53567
G=11345
H=75989
D=22456
```

That is very manageable. We can now choose E's (H>E) for each column:

```  A=33 4 777 777 888
C=88 9 666 999 444
I=44 6 222 333 222
B=55 3 555 666 777
G=11 1 333 444 555
H=77 5 999 888 999
D=22 2 444 555 666
E=06 0 018 012 013
```

10+E-H=J without a borrow, or 9+E-H=J with a borrow:

```     without borrow      with borrow
A=33 4 777 777 888   33 4 777 777 888
C=88 9 666 999 444   88 9 666 999 444
I=44 6 222 333 222   44 6 222 333 222
B=55 3 555 666 777   55 3 555 666 777
G=11 1 333 444 555   11 1 333 444 555
H=77 5 999 888 999   77 5 999 888 999
D=22 2 444 555 666   22 2 444 555 666
E=06 0 018 012 013   06 0 018 012 013
J=39 5 129 234 124   28 4 018 123 013
```

Crossing out the J's that involve conflicts:

```     without borrow      with borrow
A=33 4 777 777 888   33 4 777 777 888
C=88 9 666 999 444   88 9 666 999 444
I=44 6 222 333 222   44 6 222 333 222
B=55 3 555 666 777   55 3 555 666 777
G=11 1 333 444 555   11 1 333 444 555
H=77 5 999 888 999   77 5 999 888 999
D=22 2 444 555 666   22 2 444 555 666
E=06 0 018 012 013   06 0 018 012 013
J=x9 x 1xx 2xx 1xx   xx x xxx 12x xxx
```

We see that in the second column, where J=9, we will have a borrow, since I will be greater than F. So, that column is in conflict (since we are assuming no borrow). Similarly, both of the columns in the "with borrow" side will not cause a borrow, since F will be greater than I. So, we just have three columns. And we solve for F (the tenth letter):

```     without borrow      with borrow
A=33 4 777 777 888   33 4 777 777 888
C=88 9 666 999 444   88 9 666 999 444
I=44 6 222 333 222   44 6 222 333 222
B=55 3 555 666 777   55 3 555 666 777
G=11 1 333 444 555   11 1 333 444 555
H=77 5 999 888 999   77 5 999 888 999
D=22 2 444 555 666   22 2 444 555 666
E=06 0 018 012 013   06 0 018 012 013
J=xx x 1xx 2xx 1xx   xx x xxx xxx xxx
F=     3   5   3
```

Only one of those columns is not in conflict. And we can fill in the puzzle:

```         8
---
74)603
592
---
11
```

Normally, the smallest word arithmetics are the most difficult.