Difference between revisions of "2012 AMC 10B Problems/Problem 4"
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We know that <math>7 \equiv 1 \mod{6}</math> so therefore: | We know that <math>7 \equiv 1 \mod{6}</math> so therefore: | ||
<cmath> p + r \equiv 7 \equiv 1 \mod{6} \implies p + r \equiv 1 \mod{6} </cmath> | <cmath> p + r \equiv 7 \equiv 1 \mod{6} \implies p + r \equiv 1 \mod{6} </cmath> | ||
| − | Thus when Ringo and Paul pool their marbles, they will have <math>\boxed{\textbf{(A)}\ 1}</math> | + | Thus when Ringo and Paul pool their marbles, they will have <math>\boxed{\textbf{(A)}\ 1}</math> marble left over. |
~ herobrine-india | ~ herobrine-india | ||
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?
Solution 1
In total, there were 
marbles left from both Ringo and Paul.We know that 
. This means that there would be 
marble leftover, or 
.
Solution 2 (modulo)
Let 
be the number of marbles Ringo has and let 
be the number of marbles Paul has. we have the following equations:
![\[r \equiv 4 \mod{6}\]](http://latex.artofproblemsolving.com/b/9/c/b9c7a034974cc214bffe00ea2ddc7d64e44450e6.png)
![\[p \equiv 3 \mod{6}\]](http://latex.artofproblemsolving.com/4/6/b/46b814650decc1de3d885b245f681f988c441762.png)
Adding these equations we get:
![\[p + r \equiv 7 \mod{6}\]](http://latex.artofproblemsolving.com/9/a/5/9a57f8f669f2f2fcfe8c331d3523ad423e8a54af.png)
We know that 
so therefore:
![\[p + r \equiv 7 \equiv 1 \mod{6} \implies p + r \equiv 1 \mod{6}\]](http://latex.artofproblemsolving.com/6/9/1/691d1fe1dd73b2169a83ebf38499a10cedb84f89.png)
Thus when Ringo and Paul pool their marbles, they will have 
marble left over.
~ herobrine-india
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
