$31 \cdot 52$ satisfies the system of equations
$$
\left\{\begin{array}{l}
x+2 y+4 z=12, \\
x y+4 y z+2 x z=22, \\
x y z=6 .
\end{array}\right.
$$

The number of ordered triples $(x, y, z)$ is
(A) 0.
(B) 1.
(C) 2.
(D) 4.
(E) 6.
(27th American High School Mathematics Examination, 1976)
[Solution] Let $u=\frac{x}{2}, v=y$ and $w=2z$, then the given system of equations transforms into
$$
\left\{\begin{array}{l}
u+v+w=6, \\
uv+vw+uw=11, \\
uvw=6 .
\end{array}\right.
$$

Consider the polynomial $p(t)=(t-u)(t-v)(t-w)$, where $(u, v, w)$ is a solution to the above system of equations. If $p(t)=t^{3}-6t^{2}+11t-6$, then $(u, v, w)$ is a solution to $p(t)=0$.

On the other hand, if the solutions to $p(t)=0$ are arranged in any order as a triple, then this triple is a solution to the above system of equations.

Since $p(t)=(t-1)(t-2)(t-3)$, the six permutations of $(1,2,3)$ are solutions to the system of equations for $u, v$ and $w$. Therefore, the given system of equations for $x, y$ and $z$ has 6 solutions.
Hence, the answer is $(E)$.