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?
$\textbf{(A)}\ 24 \qquad \textbf{(B)}\ 26\qquad \textbf{(C)}\ 28 \qquad \textbf{(D)}\ 30 \qquad \textbf{(E)}\ 32$
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](https://artofproblemsolving.com/wiki/index.php/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 $\boxed{\textbf{(B)}}$.