Difference between revisions of "2013 AMC 8 Problems/Problem 1"
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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> | <math>\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5</math> | ||
| − | ==Solution== | + | ==Solution 1== |
| − | The least multiple of 6 greater than 23 is 24. So she will need to add <math>\boxed{\textbf{(A)}\ 1}</math> more model car. ~avamarora | + | The least multiple of 6 greater than 23 is 24. So she will need to add <math>24-23=\boxed{\textbf{(A)}\ 1}</math> more model car. |
| + | ~avamarora | ||
==Video Solution== | ==Video Solution== | ||
Revision as of 20:56, 1 February 2023
Contents
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
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. 
