1967 AHSME Problems/Problem 22

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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 (ProblemsAnswer KeyResources)
Preceded by
Problem 21
Followed by
Problem 23
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