Chloe charged for admission to her play on three different nights. Each night, a different number of people were in attendance, but remarkably, Chloe collected $541 each night. If the admission charges for each child and each adult were $9 and $17, respectively, how many people in total came to the three showings?

I solved this tediously by guessing and checking half a year ago in the Mathcounts Competition. I finished just the problem in time with no time to spare for checking my other answers!

Please help, I need a faster strategy to solve this.

CalculatorUser Jun 22, 2019

#1**0 **

You posted the same question about 4 weeks ago. See here for the answers:

**https://web2.0calc.com/questions/money-problem_5**

Guest Jun 22, 2019

#2**+2 **

Sorry I posted again I was confused by the explanation... I dont understand why it works.

It is locked now I couldnt reply to the answer today.

CalculatorUser
Jun 22, 2019

#3**+1 **

Here's the solution ....

Note that we should have this equation

9x + 17y = 541 which we can manipulate as follows

9x = 541 - 17y

x = 541 - 17y

________

9

x = 541 - (18 - 1)y

_____________

9

x = 541 + y 18y

_______ - ____

9 9

x = 540 + 1 + y 18y

___________ - ____

9 9

x = 540 1 + y 18y

___ + _____ - ___

9 9 9

Note that the first and last terms will always be divisible by 9

And.....the middle term will be divisible by 9 when y = 8, 17 or 26

So we have the following possibilities for x

x = 541 - 17(8) = 45

_________

9

x = 541 - 17*17 = 28

__________

9

x = 541 - 17*26 = 11

____________

9

Note if y = 35,we get the next multiple of 9...but x = -6 if this is so.... so....these are the only possible solutions

So....the solutions (x, y) = ( 11,26) (28, 17) (45, 8)

Just add these coordinates and you will get your answer, CU

CPhill Jun 22, 2019

#4**+2 **

thank you so much! that is a very beatiful solution I will make sure in the future to use the same strategy!

CalculatorUser Jun 22, 2019