Difference between revisions of "2013 AMC 8 Problems/Problem 1"

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==Solution==
 
==Solution==
In order to have her model cars in perfect, complete rows of 6, Danica must have a number of cars that is a multiple of 6. The smallest multiple of 6 which is larger than 23 is 24, so she'll need to buy <math>\boxed{\textbf{(A)}\ 1}</math> more model car.
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In order to have the model cars in complete rows of 6, Danica must have a number of cars that is a multiple of 6. The smallest multiple of 6 which is larger than 23 is 24, so she'll need to buy <math>\boxed{\textbf{(A)}\ 1}</math> more model car.
  
 
==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}}

Revision as of 16:04, 4 November 2018

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?

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

Solution

In order to have the model cars in complete rows of 6, Danica must have a number of cars that is a multiple of 6. The smallest multiple of 6 which is larger than 23 is 24, so she'll need to buy $\boxed{\textbf{(A)}\ 1}$ more model car.

See Also

2013 AMC 8 (ProblemsAnswer KeyResources)
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

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