1985 AJHSME Problems/Problem 24

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Problem

In a magic triangle, each of the six whole numbers $10-15$ is placed in one of the circles so that the sum, $S$ , of the three numbers on each side of the triangle is the same. The largest possible value for $S$ is

$\text{(A)}\ 36 \qquad \text{(B)}\ 37 \qquad \text{(C)}\ 38 \qquad \text{(D)}\ 39 \qquad \text{(E)}\ 40$

Solution

Let the number in the top circle be $a$ and then $b$ , $c$ , $d$ , $e$ , and $f$ , going in clockwise order. Then, we have

Adding these equations together, we get

$\begin{align*}3S &= (a+b+c+d+e+f)+(a+c+e) \\&= 75+(a+c+e) \\\end{align*}$

where the last step comes from the fact that since $a$ , $b$ , $c$ , $d$ , $e$ , and $f$ are the numbers $10-15$ in some order, their sum is $10+11+12+13+14+15=75$

The left hand side is divisible by $3$ and $75$ is divisible by $3$ , so $a+c+e$ must be divisible by $3$ . The largest possible value of $a+c+e$ is then $15+14+13=42$ , and the corresponding value of $S$ is $\frac{75+42}{3}=39$ , which is choice $\boxed{\text{D}}$ .

It turns out this sum is attainable if you let

1985 AJHSME (Problems • Resources)
Preceded by Problem 23	Followed by Problem 25
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1985 AJHSME Problems/Problem 24

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Problem

Solution

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