# 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

D. 21

See the Solution

### 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\$.