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Difference between revisions of "2010 AMC 12B Problems/Problem 3"

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== Solution ==
 
== Solution ==
 
We find the greatest common factor of <math>48</math> and <math>64</math> to be <math>16</math>. The number of factors of <math>16</math> is <math>5</math> which is the answer <math>(E)</math>.
 
We find the greatest common factor of <math>48</math> and <math>64</math> to be <math>16</math>. The number of factors of <math>16</math> is <math>5</math> which is the answer <math>(E)</math>.
 +
== Solution 2 ==
 +
Since the price of one ticket is a whole number, it must be a divisor of both <math>48</math> and <math>64</math>. An integer is a common divisor of two numbers if and only if it is divisible by their difference. Hence, all the possible prices are positive divisors of <math>16</math>, or <math>1</math>, <math>2</math>, <math>4</math>, <math>8</math>, and <math>16</math>, making the answer <math>5</math>, or C.
  
 
==Video Solution==
 
==Video Solution==

Revision as of 20:08, 28 August 2024

The following problem is from both the 2010 AMC 12B #3 and 2010 AMC 10B #8, so both problems redirect to this page.

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 $$48$, and a group of 10th graders buys tickets costing a total of $$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

We find the greatest common factor of $48$ and $64$ to be $16$. The number of factors of $16$ is $5$ which is the answer $(E)$.

Solution 2

Since the price of one ticket is a whole number, it must be a divisor of both $48$ and $64$. An integer is a common divisor of two numbers if and only if it is divisible by their difference. Hence, all the possible prices are positive divisors of $16$, or $1$, $2$, $4$, $8$, and $16$, making the answer $5$, or C.

Video Solution

https://youtu.be/eVp2z5Y3kUY

~Education, the Study of Everything

Video Solution

https://youtu.be/I3yihAO87CE?t=179

~IceMatrix

See also

2010 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
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 12 Problems and Solutions
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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