# Difference between revisions of "1969 AHSME Problems/Problem 24"

## Problem

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:

$\text{(A) } r>r' \text{ always}\quad \text{(B) } rr' \text{ sometimes and } rr' \text{ sometimes and } r=r' \text{ sometimes}\quad\\ \text{(E) } r=r' \text{ always}$

## Solution

The divisors are the same, so take each variable 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 $\boxed{\textbf{(E)}}$.