$29 \cdot 22$ Let $p$ be the product of any three consecutive positive odd numbers, then the largest integer that can divide all such $p$ is
(A) 15.
(B) 6.
(C) 5.
(D) 3.
(E) 1.
(19th American High School Mathematics Examination, 1968)
［Solution］Let the three positive odd numbers be $2k-1, 2k+1, 2k+3$. By the problem, $p=(2k-1)(2k+1)(2k+3)$, where $k$ is a positive integer.
(1) If 3 divides $k$, i.e., $k=3m$, then $2k+3$ is a multiple of 3;
(2) If $k=3m+1$, then $2k+1=2(3m+1)+1=6m+3$, i.e., $2k+1$ is a multiple of 3;
(3) If $k=3m+2$, then $2k-1=2(3m+2)-1=6m+3$, i.e., $2k-1$ is a multiple of 3.
In summary, $p$ always has a factor of 3. Moreover, $p$ has no other common factors, which can be seen by noting:
$$
p_{1}=1 \cdot 3 \cdot 5, \quad p_{2}=7 \cdot 9 \cdot 11
$$

Their greatest common divisor is 3. Therefore, the answer is $(D)$.