$29 \cdot 28$ When $P$ is divided by $D$, the quotient is $Q$, and the remainder is $R$, when $Q$ is divided by $D^{\prime}$, the quotient is $Q^{\prime}$, and the remainder is $R^{\prime}$, then when $P$ is divided by $D D^{\prime}$, its remainder is
(A) $R+R^{\prime} D$.
(B) $R^{\prime}+R D$.
(C) $R R^{\prime}$.
(D) $R$.
(E) $R^{\prime}$.
(18th American High School Mathematics Examination, 1967)
[Solution] Suppose we have $P=Q D+R$, and $Q=Q^{\prime} D^{\prime}+R^{\prime}$. Substituting the second equation into the first, we get
$$
P=\left(Q^{\prime} D^{\prime}+R^{\prime}\right) D+R=Q^{\prime}\left(D D^{\prime}\right)+\left(R^{\prime} D+R\right) .
$$

Since $R \leqslant D-1$, and $R^{\prime} \leqslant D^{\prime}-1$, it follows that
$$
R+R^{\prime} D \leqslant(D-1)+\left(D^{\prime}-1\right) D=D D^{\prime}-1<D D^{\prime} \text {, }
$$

thus $R+R^{\prime} D$ is the remainder.
Therefore, the answer is $(A)$.