Solve the letter values

Question

The letters have distinct integer values from 1 to 9. The totals vertically and horizontally have been given, for example: Horizontally, (D+C+A+H)=21 and vertically, (D+B+F+C)=20. Solve all the values.

 D  C  A  H  21
 B  J  F  D  18
 F  G  H  B  23
 C  A  J  E  14
20 18 12 26

Answer 1

Deduction process

1. The 3rd column is $A+F+H+J=12$, so these four letters must be either $1,2,3,6$ or $1,2,4,5$ in some order.

2. By comparing the 3rd column and 4th row, we know $C+E=F+H+2$

3. By comparing the 3rd column and 2nd row, we know $B+D=A+H+6$.

4. By comparing the 2nd column and 4th row, we know $G=E+4$.

5. Substituting the result of 4 above into the 4th column, we find $H+D+B+G=26+4=30$, so $H,D,B,G$ must be $9,8,7,6$ in some order. Putting this together with the result of 1 above, we get that

   $H=6$ and $A,F,J$ are $1,2,3$ in some order.

6. By elimination, the remaining letters $C,E$ must be

   $4,5$ in some order. By the result of 4 above, $G$ must be $8$ or $9$.

7. The result of 2 above now becomes

   $9=F+8$, i.e. $F=1$.

8. The result of 3 above now becomes $B+D=A+12$. The right-hand side must be one of $13,14,15$, while $B,D$ are two of $9,8,7$. So

   $A=3$ and $B,D$ are $8,7$ in some order. By elimination, $J=2$ and $G=9$. By the result of 4, $E=5$ and therefore $C=4$.

9. Finally, the 1st row is $D+C+A+H=21$, so

   $D=8$ and $B=7$.

Final solution

• $A=3,B=7,C=4,D=8,E=5,F=1,G=9,H=6,J=2$

• $F=1,J=2,A=3,C=4,E=5,H=6,B=7,D=8,G=9$

• 8 4 3 6 21
7 2 1 8 18
1 9 6 7 23
4 3 2 5 14
20 18 12 26

Answer 2

Well, since @Rand al’Thor has told me that it won’t harm me, here’s my answer.

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First, let’s simplify this -

      1  2  3  4
5     D  C  A  H  21 
6     B  J  F  D  18
7     F  G  H  B  23
8     C  A  J  E  14
     20 18 12 26

Now that’s done, first

Equating 8 and 2 we get -

E + 4 = G (1)

1 and 6 gives us -

C = J + 2 (2)

With 1 and 5 we have -

1 + B + F = A + H (3)

And 3 and 6 gives us -

B + D = A + H + 6 (4)

Putting (3) and (4) together proves that -

F + 7 = D (5)

2 and 6 with (5) make -

C + G + A = B + 2F + 7 (6)

Then 3 and 4 with (6) give -

F + 4 = E (7)

(1) and (7) tell us that

E can’t be 1,2,3,4,6,7,8,9 or E is 5

Using that at some of our formulae we get

D is 8, F is 1 and G is 9

Let’s fill in these numbers in the places of there respective letters. This eliminates these numbers from being the values of the other letters. Now 2 and 6 give us

B = C + A

Thus eliminating these possibilities for B -

1,2,3,4,5,8,9

2 and 5 give

J + 4 = H

1 and 3 give

C + 2 = H (8)

7 shows

H + B = 13

Adding B to (8) makes

B + C = 11

And putting (2) in there makes

B + J = 9

Now I did some assumptions (one reason why @Rand al’Thor’s answer is better than mine) and got these answers which worked

A is 3, B is 7, C is 4, H is 6 and J is 2

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Finally,

A = 3

B = 7

C = 4

D = 8

E = 5

F = 1

G = 9

H = 6

J = 2

Answer 3

Acknowledgement: this solution is motivated by Rand al'Thor's sum constraints. E.g. if a row sums to 10 and the letters are unique digits, the 4 letters must be $1234$ (dropping commas for convenience; also, the order of mapping is left undetermined in this notation).

The main feature of this solution is the use of the sum constraints as its primary deductive tool.

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Observation: C3, R4 and C2 all include A and J. Further, R4 and C2 differ by only 1 letter.

C3 (column 3) maps $AFHJ$ to one of $\{1236,1245\}$.

R4 (row 4) maps $ACEJ$ to one of $\{1238, 1247, 1256, 1346, 2345\}$.

$AJ$ is common to C3 and R4, and the remaining letters $FHCE$ are distinct. Comparing the mappings induced by C3 and R4 then, we cannot have $AFHJ = 1236, ACEJ = 1238$ because then $3$ would be assigned to 2 different letters.

A quick check of the 2x5 possibilities leaves us with 4 cases, denoted $Kn$:

$$ K1: AFHJ=1236, ACEJ=1247 \implies AJ=12 \\ K2: AFHJ=1236, ACEJ=2345 \implies AJ=23 \\ K3: AFHJ=1245, ACEJ=1238 \implies AJ=12 \\ K4: AFHJ=1245, ACEJ=1346 \implies AJ=14 $$

Now consider C2. We need to find mappings that preserve the $AJ$ mappings from before, and have a third common digit between R4 and C2 (because $C$ is also common to those two quartets), while all the other digits must be distinct.

So C2's $ACGJ$ must map to one of $\{1278 (K1,K3), 1467 (K4), 2349 (K2), 2358 (K2)\}$.

Say $ACGJ = 1278$ with case $K3$. We have $AJ=12$, and $C$ is remaining digit common to R4 and C2, so $C=8$ and hence $E=3,G=7$, leaving $FH=45$. Plugging these values into R3, though, gives us $B=3=G$, which violates the unique letter-mapping constraint.

Repeating this exercise eliminates all cases except the two $K2$ possibilities. In each case, we have mapped 8 of the 9 digits, so $D$ can only be the respective case's remaining digit. R2 eliminates the case with $B=9$, so the only possibility left is: $$AJ=23, C=4, E=5, G=9, FH=16, B=7, D=8$$

Use R1 to disambiguate: $A+H=9$, so the final solution is: $$A=3, J=2, C=4, E=5, G=9, H=6, F=1, B=7, D=8$$

In alphabetical order, $(A,B,C,D,E,F,G,H,J) = (3,7,4,8,5,1,9,6,2)$.

This agrees with the solutions of both Rand al'Thor and Napoleon of Puzzling. QED.