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

## See Also

 1969 AHSC (Problems • Answer Key • Resources) Preceded byProblem 23 Followed byProblem 25 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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