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

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?

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

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)
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.

@@ Line 9: / Line 9: @@
 ==Solution 2==
-<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>.
+x 4 = 24, which is 1 more than 23. So, the answer is <math>\boxed{\textbf{(A)}\ 1}</math>.
-~megaboy6679
 ==Video Solution==

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