1963 AHSME Problems/Problem 23

Problem

A gives $B$ as many cents as $B$ has and $C$ as many cents as $C$ has. Similarly, $B$ then gives $A$ and $C$ as many cents as each then has. $C$ , similarly, then gives $A$ and $B$ as many cents as each then has. If each finally has $16$ cents, with how many cents does $A$ start?

$\textbf{(A)}\ 24 \qquad \textbf{(B)}\ 26\qquad \textbf{(C)}\ 28 \qquad \textbf{(D)}\ 30 \qquad \textbf{(E)}\ 32$

Solution

Let $a$ be number of cents $A$ originally had, $b$ be number of cents $B$ originally had, and $c$ be number of cents $C$ originally had.

After $A$ gave his money away, $A$ has $a-b-c$ cents, $B$ has $2b$ cents, and $C$ has $2c$ cents.

After $B$ gave his money away, $A$ has $2a-2b-2c$ cents, $B$ has $-a+3b-c$ cents, and $C$ has $4c$ cents.

After $C$ gave his money away, $A$ has $4a-4b-4c$ cents, $B$ has $-2a+6b-2c$ cents, and $C$ has $-a-b+7c$ cents.

Since all of them have $16$ cents in the end, we can write a system of equations. $4a-4b-4c=16$ $-2a+6b-2c=16$ $-a-b+7c=16$ Note that adding the three equation yields $a+b+c=48$ , so $4a+4b+4c=192$ . Therefore, $8a=208$ , so $a = 26$ . Solving for $a$ can also be done traditionally.

Thus, $A$ started out with $26$ cents, which is answer choice $\boxed{\textbf{(B)}}$ .

Solution 2

Working backward, we know that people not giving away money on the previous turn now have twice what they previously had. Using the fact that the sum of their

1963 AHSC (Problems • Answer Key • Resources)
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1963 AHSME Problems/Problem 23

Contents

Problem

Solution

Solution 2

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