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

(Created page with "== Problem == For natural numbers, when <math>P</math> is divided by <math>D</math>, the quotient is <math>Q</math> and the remainder is <math>R</math>. When <math>Q</math> is d...")
 
Line 9: Line 9:
  
 
== Solution ==
 
== Solution ==
<math>\fbox{A}</math>
+
We are given <math>P = QD + R</math> and <math>Q = D'Q' + R'</math>.
 +
 
 +
Plugging the second equation into the first yields:
 +
 
 +
<math>P = (D'Q' + R')D + R</math>
 +
<math>P = (DD')Q' + (R'D + R)</math>
 +
 
 +
If we divide <math>P</math> by <math>DD'</math>, the quotient would be <math>Q'</math>, and the remainder would be <math>R'D + R</math>, which is option <math>\fbox{A}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 19:15, 12 July 2019

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
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. AMC logo.png

Invalid username
Login to AoPS