29.44 The expression $21 x^{2}+a x+21$ can be factored into two linear factors, where $a$ is a positive integer. Then $a$ is
(A) any odd number.
(B) some odd number.
(C) any even number.
(D) some even number.
(E) zero.
(Shenzhen, Guangdong Province, China Junior High School Mathematics Competition, 1988)
[Solution] Let $21 x^{2}+a x+21=(m x+n)(p x+q)$, then $21 x^{2}+a \cdot x+21=m p x^{2}+(m q+n p) x+n q$,

So $\left\{\begin{array}{l}m p=21 \\ m q+n p=a \\ n q=21\end{array}\right.$
From the above (1) and (3) equations, we know that $m, n, p, q$ are all odd numbers.
Then $m q+n p$ must be an even number, i.e., $a$ is some even number.
Therefore, the answer is $(D)$.