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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. <math>7/6</math>=1R1. <math>\text{This means that that there is}</math>                <math> \boxed{1}</math> <math>\text{marbles left}</math> or
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== Problem 4 ==
  
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['''A''']
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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?
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<math> \textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5 </math>
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== Solution ==
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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> .
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==See Also==
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{{AMC10 box|year=2012|ab=B|num-b=3|num-a=5}}
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{{MAA Notice}}

Revision as of 21:09, 8 February 2014

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 $\boxed{A}$ .

See Also

2012 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
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
All AMC 10 Problems and Solutions

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