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

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==Problem==
 
==Problem==
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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==
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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>
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==Solution 1==
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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.
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~avamarora
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==Solution 2==
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<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>.
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~megaboy6679
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==Video Solution==
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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}}

Revision as of 22:14, 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?

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

Solution 1

The least multiple of 6 greater than 23 is 24. So she will need to add $24-23=\boxed{\textbf{(A)}\ 1}$ more model car. ~avamarora

Solution 2

\[23\equiv5\equiv-1\pmod6\]For the number of cars to be a multiple of $6$, we have to increase the number of cars by $\boxed{\textbf{(A)}~1}$.

~megaboy6679

Video Solution

https://youtu.be/HcWVIEnH0vs ~savannahsolver

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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