# Difference between revisions of "1967 AHSME Problems/Problem 22"

## Problem

For natural numbers, when $P$ is divided by $D$, the quotient is $Q$ and the remainder is $R$. When $Q$ is divided by $D'$, the quotient is $Q'$ and the remainder is $R'$. Then, when $P$ is divided by $DD'$, the remainder is:

$\textbf{(A)}\ R+R'D\qquad \textbf{(B)}\ R'+RD\qquad \textbf{(C)}\ RR'\qquad \textbf{(D)}\ R\qquad \textbf{(E)}\ R'$

## Solution

We are given $P = QD + R$ and $Q = D'Q' + R'$.

Plugging the second equation into the first yields:

$P = (D'Q' + R')D + R$

$P = (DD')Q' + (R'D + R)$

If we divide $P$ by $DD'$, the quotient would be $Q'$, and the remainder would be $R'D + R$, which is option $\fbox{A}$.

## See also

 1967 AHSME (Problems • Answer Key • Resources) Preceded byProblem 21 Followed byProblem 23 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 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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