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Difference between revisions of "2010 AMC 10B Problems/Problem 8"

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==Solution 2==
 
==Solution 2==
  
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The difference between 48 and 64 is 16. The number 16 has five factors: 1, 2, 4, 8, and 16. These are all possible ticket prices because the 10th graders bought a whole number <math>x</math> more tickets than the 9th graders. Therefore, there are <math>\boxed{\textbf{(E)}\ 5}</math> possibilities for the ticket price.
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The difference between 48 and 64 is 16. The number 16 has five factors: 1, 2, 4, 8, and 16. These are all possible ticket prices because the 10th graders bought a whole number <math>x</math> more tickets than the 9th graders (so <math>x</math> must evenly divide into 16). Therefore, there are <math>\boxed{\textbf{(E)}\ 5}</math> possibilities for the ticket price.
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2010|ab=B|num-b=7|num-a=9}}
 
{{AMC10 box|year=2010|ab=B|num-b=7|num-a=9}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 18:00, 9 October 2017

Problem

A ticket to a school play cost $x$ dollars, where $x$ is a whole number. A group of 9th graders buys tickets costing a total of $\textdollar 48$, and a group of 10th graders buys tickets costing a total of $\textdollar 64$. How many values for $x$ are possible?

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

Solution 1

We see how many common integer factors $48$ and $64$ share. Of the factors of $48$ - $1, 2, 3, 4, 6, 8, 12, 16, 24, 48$; only $1, 2, 4, 8,$ and $16$ are factors of $64$. So there are $\boxed{\textbf{(E)}\ 5}$ possibilities for the ticket price.

Solution 2

The difference between 48 and 64 is 16. The number 16 has five factors: 1, 2, 4, 8, and 16. These are all possible ticket prices because the 10th graders bought a whole number $x$ more tickets than the 9th graders (so $x$ must evenly divide into 16). Therefore, there are $\boxed{\textbf{(E)}\ 5}$ possibilities for the ticket price.

See Also

2010 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
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 AMC 10 Problems and Solutions

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