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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

See Also

1963 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40
All AHSME Problems and Solutions

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