$36 \cdot 58 A$ gives $B$ an amount of money equal to the amount of money $B$ has, $A$ gives $C$ an amount of money equal to the amount of money $C$ has, similarly, $B$ gives $A$ and $C$ an amount of money equal to the amount of money each of them has, $C$ gives $A$ and $B$ an amount of money equal to the amount of money each of them has, if in the end each of them has 16 dollars, then the initial amount of money $A$ has is
(A) 24.
(B) 26.
(C) 28.
(D) 30.
(E) 32.
(14th American High School Mathematics Examination, 1963)
[Solution] Let the initial amounts of money that $A, B, C$ have be $a, b, c$ respectively. Based on this, we can construct the following table:
\begin{tabular}{c|c|c|c}
\hline & $A$ & $B$ & $C$ \\
\hline Start & $a$ & $b$ & $c$ \\
\hline After the first round & $a-b-c$ & $2 b$ & $2 c$ \\
\hline After the second round & $2(a-b-c)$ & \begin{tabular}{l}
$2 b-(a-b-c)-2 c$ \\
$=3 b-a-c$
\end{tabular} & $4 c$ \\
\hline After the third round & $4(a-b-c)$ & $2(3 b-a-c)$ & \begin{tabular}{r}
$4 c-2(a-b-c)$ \\
$-(3 b-a-c)$ \\
\\
$7 c-a-b$
\end{tabular} \\
\hline
\end{tabular}

Then we have the system of equations:
\[
\left\{\begin{array}{l}
4(a-b-c)=16, \\
2(3 b-a-c)=16, \\
7 c-a-b=16 .
\end{array}\right.
\]
Solving this system, we get:
\[
\left\{\begin{array}{l}
a=26, \\
b=14, \\
c=8 .
\end{array}\right.
\]
Therefore, the answer is $(B)$.