2010 AMC 12B Problems/Problem 3

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 E. - By Monkey_King

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 (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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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