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 ==
  
['''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 leftover?
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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 1 ==
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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 leftover, or <math>\boxed{A}</math>.
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== Solution 2 (modulo) ==
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Let <math>r</math> be the number of marbles Ringo has and let <math>p</math> be the number of marbles Paul has. we have the following equations:
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<cmath> r \equiv 4 \mod{6} </cmath>
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<cmath> p \equiv 3 \mod{6} </cmath>
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Adding these equations we get:
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<cmath> p + r \equiv 7 \mod{6} </cmath>
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We know that <math>7 \equiv 1 \mod{6}</math> so therefore:
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<cmath> p + r \equiv 7 \equiv 1 \mod{6} \implies p + r \equiv 1 \mod{6} </cmath>
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Thus when Ringo and Paul pool their marbles, they will have <math>\boxed{\textbf{(A)}\ 1}</math> marble left over.
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~ herobrine-india
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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}}

Latest revision as of 18:07, 13 July 2021

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 leftover?

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


Solution 1

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 leftover, or $\boxed{A}$.

Solution 2 (modulo)

Let $r$ be the number of marbles Ringo has and let $p$ be the number of marbles Paul has. we have the following equations: \[r \equiv 4 \mod{6}\] \[p \equiv 3 \mod{6}\] Adding these equations we get: \[p + r \equiv 7 \mod{6}\] We know that $7 \equiv 1 \mod{6}$ so therefore: \[p + r \equiv 7 \equiv 1 \mod{6} \implies p + r \equiv 1 \mod{6}\] Thus when Ringo and Paul pool their marbles, they will have $\boxed{\textbf{(A)}\ 1}$ marble left over.

~ herobrine-india

See Also

2012 AMC 10B (ProblemsAnswer KeyResources)
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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