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

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 money is $48$ cents, we can work backward $16,16,16$ $8,8,32$ $4,28,16$ $26,14,8$ Thus at the beginning $A$ has $26\boxed{B}$ cents.

~ Nafer

1963 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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1963 AHSME Problems/Problem 23

Contents

Problem

Solution

Solution 2

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