When the natural numbers $P$ and $P'$, with $P>P'$, are divided by the natural number $D$, the remainders are $R$ and $R'$, respectively. When $PP'$ and $RR'$ are divided by $D$, the remainders are $r$ and $r'$, respectively. Then:
$\text{(A) } r>r' \text{  always}\quad \text{(B) } rr' \text{  sometimes and } rr' \text{  sometimes and } r=r' \text{  sometimes}\quad\\ \text{(E) } r=r' \text{  always}$
The divisors are the same, so take each variable [modulo](https://artofproblemsolving.com/wiki/index.php/Modulo) $D$.
\[P \equiv R \pmod{D}\]
\[P' \equiv R’ \pmod{D}\]
That means
\[PP’ \equiv RR’ \pmod{D}\]
Thus, $PP’$ and $RR’$ have the same remainder when divided by $D$, so the answer is $\boxed{\textbf{(E)}}$.