2013 AMC 8 Problems/Problem 1

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

6 x 4 = 24, which is 1 more than 23. So, the answer is $\boxed{\textbf{(A)}\ 1}$ .

Video Solution

https://youtu.be/HcWVIEnH0vs ~savannahsolver

2013 AMC 8 (Problems • Answer Key • Resources)
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All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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2013 AMC 8 Problems/Problem 1

Contents

Problem

Solution 1

Solution 2

Video Solution

See Also