The price of concert tickets was $25 for local students, $50 for local nonstudents, and $120 for foreigners.
There are 100 tickets sold for a total of $4000.
It is also known that at least one tick of each type was sold.
How many tickets of each type were sold?

- ganeshie8

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- ganeshie8

@Loser66
give it a try :)

- Loser66

it looks liked algebra with the equations
25x + 50y + 120 z = 4000
x+y+z = 100
but they are have 2 equations with 3 unknown.

- ganeshie8

yes, the tricky part is that it must be solved over "integers" as you cannot buy tickets in fractional amounts..

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## More answers

- ganeshie8

\(x,y,z\in \mathbb{N}\)

- lochana

x = 68
y = 22
z = 10

- lochana

68*25 = 1700
22*50 = 1100
120*10 = 1200
1700+1100+1200 = 2800+1200 = 4000
@ganeshie8 am i correct?

- Loser66

I think to solve it, we need express 1 in terms of 5, 10 and 24, right?

- ganeshie8

Yep! (68, 22, 10) is one solution,
how do we know thats the only solution ?

- ganeshie8

@Loser66
that should work, you could also make the work simple by eliminating a variable before doing that..

- lochana

yes. that's what i was about to ask.
however,
you need $120 to be 0 in its 10^2 base. like 1200 , 2400 etc. otherwise. we can't combine multiplication of $120 with multiplication of $50 and $25 to be made $4000. So I started with 10 tickets of 120$. and then got two equations with two variables.

- ParthKohli

\[x = 100 - y - z \]\[5x + 10y + 24 z = 800\]\[5(100 - y - z) + 10y + 24 z = 800\]\[5y + 19z = 300\]This is only possible when \(z \) is a multiple of five. Let's plug that into the original equation and try stuff.\[5x + 10y + 24z = 800\]Case I: \(z = 0\), meaning \(x + y = 100\). \(5x + 1000 - 10x = 800\Rightarrow x = 40, y = 60\). Found another one.

- ParthKohli

\[z = 5 \Rightarrow x + y = 95, ~5x+10y + 120 = 800 \Rightarrow x = 54,~ y = 41\]

- lochana

@ParthKohli can we take z,x or y =0 situations?, if that's the case we already have 3 answers.

- ParthKohli

\[z = 10 \Rightarrow x+y=90, ~5x+10y + 240 = 800 \Rightarrow \text{@lochana's solution}\]

- ParthKohli

\[z=15\Rightarrow x + y = 85, 5x + 10y + 360 = 800\Rightarrow x = 82, y = 3 \]

- ganeshie8

That's nice! yeah z must be a multiple of 5
let me add \(x,y,z > 0\) to the main problem..

- lochana

wow. that's a good one.

- ParthKohli

We can do this till \(z = 30\).

- ParthKohli

z=20 doesn't work though haha

- ganeshie8

there are exactly 3 solutions...

- ParthKohli

yes, that's right. at \(z = 5, 10, 15\). afterwards, it's just negative.

- ganeshie8

Thats it !
@Loser66 are you also getting the same by solving the problem using diophantine method ?

- ganeshie8

\(x = 100 - y - z\tag{1} \)
\(25x + 50y + 120 z = 4000\)
\(\implies 5x + 10y + 24 z = 800\tag{2}\)
eliminating \(x\) gives
\(5(100 - y - z) + 10y + 24 z = 800\)
\(5y + 19z = 300\tag{3}\)
This is a linear diophantine equation in two variables.

- ParthKohli

hmm, apparently there's no need to eliminate to see that z is a multiple of five. >_<
the solution could have been made shorter.

- ganeshie8

Haha right, since \((24, 5)=1\), it follows that \(z\mid 5\).
But two variables are easy to deal with using the diophantine method..

- lochana

@ganeshie8 how did you say that it has only 3 solutions? any method?

- ganeshie8

solving \(5y + 19z = 300\) over integers :
step1 : finding gcd of 5 and 19
19 = 5*3 + 4
5 = 4*1 + 1
step2 : expressing gcd as combination of 5 and 19
1 = 5 - 4*1 = 5 - (19 - 5*3)*1 = 5(1+3) - 19*1 = 5(4) + 19(-1)
step3 : particular solution
1 = 5(4) + 19(-1)
multiplying 300 through out gives
300 = 5(1200) + 19(-300)
so (1200, -300) is one particular solution to the equation 5y+19z = 300
step4 : null solution
(-19, 5) is a solution to the equation 5y + 19z = 0
step5 : all the solutions
(y, z) = (1200, -300) + t(-19, 5)

- ganeshie8

y = 1200 - 19t
z = -300 + 5t
since we want the solutions only in positive integers :
1200 - 19t > 0
-300 + 5t > 0
solving gives \(t \in \{61, 62, 63\}\)
we can plugin these \(t\) values to get the corresponding \(y, z\) values
@Loser66

- ganeshie8

@lochana thats the general mehtod to solve any linear diophantine equation...

- Loser66

Question: is it not that if 5y +19 z = 300, then \(5y\equiv 300\equiv 15 (mod 19)\) and (5,9) =1, we can have \(y\equiv 3 (mod 19)\)

- ganeshie8

thats correct

- ganeshie8

Notice that \(y = 1200 - 19t\) is exactly same as saying \(y\equiv 3\pmod{19}\)

- ganeshie8

your method is just an alternative way, which avoids euclid gcd algorithm, to find a paritcular solution.

- ganeshie8

However, you need to solve "two" different congruences to find a particular solution, one for each variable :
\(5y +19 z = 300\)
\[5y\equiv 300\pmod{19}\tag{1} \]\[19z\equiv 300\pmod{5}\tag{2}\]

- Loser66

Another question, when we get y, like above, can we put it back to original one to eliminate y, then we have just 2 unknowns for 2 equations. Then we again, use substitute to find x, z.

- Loser66

I meant: if we let minimum y = 21, put back to get
x + z = 79
25x +120z = 2950

- ganeshie8

you could do that, but that wont reduce the amount of work, it only increases..

- Loser66

we add them together: 26x +121z = 3029
then again, use \(26x \equiv 3029 (mod 121)\) to solve for x

- ganeshie8

how did u get y = 21 ?

- Loser66

Actually, we don't need to reduce work to get the answer. As long as we can apply the lecture to solve the problem, we are ok?

- Loser66

oh, I am tired, it is not 21, it is 22 .

- ganeshie8

we are not ok,
y can take infinitely many integer values.
we're not achieving anything useful by plugging just one value in the equation...

- Loser66

Not that, if we take value of x, y, z too large, the solution is not ok since they are restricted by the amount of money 4000. We need be back to original problem to solve, right?

- ganeshie8

how is this any different from guessing ?

- Loser66

It is not different but "what is the goal of lecture?"

- ganeshie8

you want to try all y values and pick the ones that work
that doesn't look like a better method

- ganeshie8

what lecture ?

- Loser66

diophantine equation solving.

- ganeshie8

diophanitne mehtod is like an algorithm
no guessing is necessary at any step of the procedure..

- Loser66

Like what lochana posted, at the first look, I thought that is guessing,
then parkholi used algebra to solve.
Surely we can apply any method to solve but that is not good for me.

- ganeshie8

diophantine method gives you directly the "y" values that work,
we don't need to verify a bunch of y values...

- Loser66

no, the restrain x+y +z=100, doesn't allow us to have many option to verify.

- ganeshie8

what would you do if x + y + z = 100000000 ?
it doesn't matter if they are few or plenty, you're still verifying and rejecting values that don't work

- Loser66

Ok.

- Willie579

I'm in 7th grade and I'm gonna try to figure this out. XD

- ganeshie8

good luck! do let us know your answer when you have it :)

- Willie579

@ganeshie8 I'm so sorry, I forgot all about this. =(

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