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:

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

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 $\boxed{\textbf{(E)}}$ .

1969 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
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1969 AHSME Problems/Problem 24

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