Ticket - PS

9X + 5Y = 927 (927 is divisible by 9)

Y should therefore be a multiple of 9 too (or) 5Y should be a multiple of 45.

5Y = 45A, and A can take value from '0' to (927/45) ~ 20

0 to 20 is 21 different cases

karthikpandian19 wrote:A high school sells tickets to a basketball game. Adult tickets cost $9 and student tickets cost $5. If n tickets are sold for a total revenue of $927, how many possible values are there for n?

A. 18
B. 19
C. 20
D. 21
E. 22

Let a = the number of adult tickets and s = the number of student tickets:
9a + 5s = 927.

An integer is a multiple of 9 if the sum of its digits is a multiple of 9.
Since 9+2+7 = 18, 927 is a multiple of 9.
Thus, 9a + 5s = multiple of 9.

9a is a multiple of 9.
Thus:
(multiple of 9) + 5s = (multiple of 9)
5s = (multiple of 9) - (multiple of 9).
The difference of two multiples of 9 must also be a multiple of 9.
Thus, the revenue from the student tickets is a multiple of 9.
Let the revenue from the student tickets = 9(5s).

Since every term in 9a + 9(5s) = 927 is a multiple of 9, we can divide each term by 9:
a + 5s = 103.
The total number of tickets = a+s.
Since a and s must be nonnegative integers, the smallest possible value of 5s = 5*0 = 0 and the largest possible value of 5s = 5*20 = 100.
Thus, s can be any integer 0 through 20, yielding 21 possible combinations for a+s.

The correct answer is D.