20. The length and width of a rectangle are both positive integers, the area is $A$ square centimeters, and the perimeter is $P$ centimeters. Then the number that cannot be $A+P$ is ( ).
(A) 100
(B) 102
(C) 104
(D) 106
(E) 108
20. B.

Let the length of a rectangle be $a$, and the width be $b$, where $a, b \in \mathbf{Z}_{+}, a > b$.
$$
\begin{array}{l}
\text { Then } A+P=ab+2a+2b \\
\Rightarrow A+P+4=ab+2a+2b+4 \\
\quad=(a+2)(b+2),
\end{array}
$$

That is, $A+P+4$ can be factored into the product of two positive integers greater than 2.

But $102+4=106=2 \times 53$, so $A+P$ cannot be 102.