Starting from the bottom of the diagram, we first determine the value of $F$. From the conditions given, either $F-7=3$ or $7-F=3$. Since $F$ must be one of $1,2,4,5,6$, and 8 , then $F=4$, giving

$$
\begin{array}{llll}
A & 10 & B & C
\end{array}
$$

![](https://cdn.mathpix.com/cropped/2024_04_20_faa9db06d997f2a5b9b1g-200.jpg?height=49&width=157&top_left_y=878&top_left_x=534)

$\begin{array}{ll}7 & 4 \\ & \\ & 3\end{array}$

3

Similarly, we can determine that $E=5$ and $D=2$, giving

## $\begin{array}{llll}A & 10 & B & C\end{array}$

![](https://cdn.mathpix.com/cropped/2024_04_20_faa9db06d997f2a5b9b1g-200.jpg?height=55&width=155&top_left_y=1176&top_left_x=535)

| 7 | 4 |
| :--- | :--- |

3

and then $A=8, B=1$, and $C=6$. (We notice that this does use each of the six possibilities exactly once.)

Thus, $A+C=14$.

ANSWER: (E)
In the diagram, the numbers $1,2,4,5,6$, and 8 are substituted, in some order, for the letters $A, B, C, D, E$, and $F$, so that the number

$\begin{array}{llll}A & 10 & B & C\end{array}$ between and below two numbers is the positive difference between those two numbers. For example, the 7 in the third row is the D $9 \quad E$ $7 \quad F$

3 positive difference between $D$ and 9 . Thus $D=2$ because $9-2=7$. The value of $A+C$ is
(A) 7
(B) 12
(C) 13
(D) 10
(E) 14