Eight distinct integer values between $1$ and $8$ satisfy six simple conditions ($D+E=F$, etc). Find which value is which.

Question

I have 8 variables; $A$, $B$, $C$, $D$, $E$, $F$, $G$, and $H$.

What are they all equal to?

Rules:

1. All the variables are equal to integer values between one and eight.

2. None of the variables are equal to each other.

3. $D + E = F$

4. $B + C = D$

5. $B + H = A$

6. $C + G = F$

7. $E + G = H$

8. $B + E = G$

I know that $1$ and $2$ both must be either $B$, $C$, or $E$ since there is no $0$ so $1$ must be a letter that does not have an equation answer. For $2$ I know that there is no equation that is the a variable plus itself.

Answer 1

As given in the post ${B,C,E}$ contains the numbers $1$ and $2$. Similarly ${A,F}$ contains 8.

The equations give $A=2(B+E),H=B+2E,G=B+E$. $A$ must therefore be either $6$ or $8$.

If $A=6$, then $B=1,E=2, F=8,G=3,H=5$ or $B=2,E=1,G=3,F=8,H=4$.

If $A=8$, then $B=1,C=2,E=3,G=4,H=7$ or $B=3,C=2,E=1,G=4,H=5$.

The actual solution is $A=6,B=2,C=5,D=7,E=1,F=8,G=3,H=4$.

Answer 2

Some hints:

Since $8$ cannot occur on the left hand side of an equation, we know that either $A=8$ or $F=8$.

From equations $1$ and $2$ we know that $B+C+E=F$. So $F=6,7$ or $8$.

From equations $5,7,8$ we know that $A=B+H = B+E+G=2G$. Since $G \ge 3$ we have either $G=3, A=6$ or $G=4, A=8$.

Suppose $G=3$. Then $A=6$ and so $F=8$. From equation $8$ we know that $B$ and $E$ are $1$ and $2$ (either way round), so $C=5$ (because $B+C+E=F=8$). Since $E+G=H$ we know that $E$ cannot be $2$ (otherwise $H=C=5$) so $E=1$ and $B=2$.

This should be enough for you to find one solution.

You should also consider the case "Suppose $G=4$ ..." to see if there are any other solutions.