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Difference between revisions of "2012 AMC 10B Problems/Problem 4"

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In total, there were <math>3+4=7</math> marbles left from both Ringo and Paul.We know that  <math>7 \equiv 1 \pmod{6}</math>. This means that there would be <math>1</math> marble left over, or <math>\boxed{A}</math> .
 
In total, there were <math>3+4=7</math> marbles left from both Ringo and Paul.We know that  <math>7 \equiv 1 \pmod{6}</math>. This means that there would be <math>1</math> marble left over, or <math>\boxed{A}</math> .
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Revision as of 12:14, 4 July 2013

Problem 4

When Ringo places his marbles into bags with 6 marbles per bag, he has 4 marbles left over. When Paul does the same with his marbles, he has 3 marbles left over. Ringo and Paul pool their marbles and place them into as many bags as possible, with 6 marbles per bag. How many marbles will be left over?

$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$


Solution

In total, there were $3+4=7$ marbles left from both Ringo and Paul.We know that $7 \equiv 1 \pmod{6}$. This means that there would be $1$ marble left over, or $\boxed{A}$ . The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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