2003 CEMC Pascal Problems/Problem 7

Problem

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 between and below two numbers is the positive difference between those two numbers. For example, the $7$ in the third row is the positive difference between $D$ and $9$ . Thus D = $2$ because $9 - 2 = 7$ . The value of $A + C$ is

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$\text{ (A) }\ 7 \qquad\text{ (B) }\ 12 \qquad\text{ (C) }\ 13 \qquad\text{ (D) }\ 10 \qquad\text{ (E) }\ 14$

Solution

We can replace $D$ with $2$ in the image:

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Since $A$ and $10$ are above $2$ , we have:

$A - 10 = 2$ or $10 - A = 2$

This gives:

$A = 12$ or $A = 8$

However, $A = 12$ is not one of the possible values listed in the original problem, so $A = 8$ .

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We can use a similar process for $F$ (since $9$ is not possible for $F$ ) to get $F = 4$ . $F = 4$ then allows us to get $E = 5$ .

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$B$ and $10$ are above $9$ , so $B = 1$ . Since $1$ , $2$ , $4$ , $5$ , and $8$ have all been used, $C = 6$ .

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We can now use $A = 8$ and $C = 6$ to get $A + C = 8 + 6 = \boxed {\textbf {(E) } 14}$ .

~anabel.disher

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2003 CEMC Pascal Problems/Problem 7

Problem

Solution