Competition problem — Combinatorics

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Combinatorics Difficulty 3.2 AMC 10/12 Find the answer

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?

Pick one

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 $\boxed{\textbf{(B)}}$ .

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.