# Problem of the Day

A high school sells tickets to a basketball game. Adult tickets cost $9 and student tickets cost $5. If tickets are sold for a total revenue of $927, how many possible values are there for ?

A. 18

B. 19

C. 20

D. 21

E. 22

### Solution

[latexpage]

From the given information we can set up the following Â equation to model the revenue, where $x$ represents the number of student tickets sold and $y$ represents the number of adult tickets sold: $5x+9y=927$. First, note that $x$ and $y$ are both non-negative integers. Next, we can see that 927 is a multiple of 9 because its digits add to 18, and $9y$ is clearly a multiple of 9. This implies that $5x$ is also a multiple of 9, and since 5 does not contain any factors of 9, $x$ itself must be a multiple of 9. That is, all possible values of $x$ are of the form $x=9t$ for some non-negative integer, $t$. Plugging this in to the original equation: $5(9t)+9y=927$. Solve this for $y$:

$45t+9y=927$

$5t+y=103$

$y=103-5t$

Values of $y$ must be non-negative integers, so it must be true that $103-5t >=0$. Solving this inequality gives us $t<=20.6$. We initially defined $t$ to be a non-negative integer, so $0<= t <= 20$, which means there are 21 possible values for $t$. To show that this means that there are 21 possible values of $n$, note that $n=x+y$, so $n=9t+103-5t$ or $n=103+4t$. Clearly each of the possible 21 values for $t$ will produce a distinct value for $n$.

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