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>24-23=\boxed{\textbf{(A)}\ 1}</math> more model car. | |
| + | ~avamarora | ||
| + | |||
| + | ==Solution 2== | ||
| + | 6 x 4 = 24, which is 1 more than 23. So, the answer is <math>\boxed{\textbf{(A)}\ 1}</math>. | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/HcWVIEnH0vs ~savannahsolver | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2013|before=First Problem|num-a=2}} | {{AMC8 box|year=2013|before=First Problem|num-a=2}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 15:14, 16 August 2024
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
6 x 4 = 24, which is 1 more than 23. So, the answer is 
.
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. 
