Difference between revisions of "2013 AMC 8 Problems/Problem 1"
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==Solution 2== | ==Solution 2== | ||
<cmath>23\equiv5\equiv-1\pmod6</cmath>For the number of cars to be a multiple of <math>6</math>, we have to increase the number of cars by <math>\boxed{\textbf{(A)}~1}</math>. | <cmath>23\equiv5\equiv-1\pmod6</cmath>For the number of cars to be a multiple of <math>6</math>, we have to increase the number of cars by <math>\boxed{\textbf{(A)}~1}</math>. | ||
| + | |||
| + | ~megaboy6679 | ||
==Video Solution== | ==Video Solution== | ||
Revision as of 21:14, 1 February 2023
Problem
Danica wants to arrange her model cars in rows with exactly 6 cars in each row. She now has 23 model cars. What is the smallest number of additional cars she must buy in order to be able to arrange all her cars this way?
Solution 1
The least multiple of 6 greater than 23 is 24. So she will need to add 
more model car.
~avamarora
Solution 2
![\[23\equiv5\equiv-1\pmod6\]](http://latex.artofproblemsolving.com/2/e/3/2e3b6a136ee3dde22f5d3db6ade156fb9bfca353.png)
For the number of cars to be a multiple of 
, we have to increase the number of cars by 
.
~megaboy6679
Video Solution
https://youtu.be/HcWVIEnH0vs ~savannahsolver
See Also
| 2013 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by First Problem |
Followed by Problem 2 | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
