32.52 Given the polynomial $P(x)=a_{0} x^{n}+a_{1} x^{n-1}+\cdots+a_{n-1} x+a_{n}$, where $n$ is a non-negative integer, $a_{0}$ is a positive integer, $a_{1}, a_{2}, \cdots, a_{n}$ are integers, and it satisfies $n+$ $\left|a_{0}\right|+\left|a_{1}\right|+\cdots+\left|a_{n}\right|=3$. Then the number of such polynomials is
(A) 4.
(B) 5.
(C) 6.
(D) Does not exist.
(China Sichuan Province High School Mathematics Competition, 1993)
[Solution] Since $a_{0}$ is a positive integer, $n,\left|a_{1}\right|,\left|a_{2}\right|, \cdots,\left|a_{n}\right|$ are non-negative integers, then $n+a_{0}+\left|a_{1}\right|+\left|a_{2}\right|+\cdots+\left|a_{n}\right|=3$ has the following 5 sets of solutions:
$$
\begin{array}{l}
\left\{\begin{array} { l } 
{ n = 0 , } \\
{ a _ { 0 } = 3 ; }
\end{array} \quad \left\{\begin{array} { l } 
{ n = 1 , } \\
{ a _ { 0 } = 2 ; }
\end{array} \quad \left\{\begin{array}{l}
n=1, \\
a_{0}=a_{1}=1 ;
\end{array}\right.\right.\right. \\
\left\{\begin{array} { l } 
{ n = 1 , } \\
{ a _ { 0 } = 1 , } \\
{ a _ { 1 } = - 1 ; }
\end{array} \quad \left\{\begin{array}{l}
n=2, \\
a_{0}=1 .
\end{array}\right.\right. \\
\end{array}
$$

Thus, the required polynomials are $3 x^{0}, 2 x, x+1, x-1, x^{2}$, a total of 5. Therefore, the answer is (B).