# What is a systematic method to solve this Diophantine equation?

1. Oct 2, 2012

### the_Doctor111

1. The problem statement, all variables and given/known data

Suppose you spend $12.30 on chocolate bars and chips. If chocolate bars cost$1.20 and each bag of chips cost \$2.50, how many bags of chips did you buy?

*Both the number of chocolate bars and bags of chips must be positive.

2. Relevant equations

12.30 = 1.20x + 2.50y

3. The attempt at a solution

What is the method i go about solving this problem besides guess and check. I assume it has something to do with the euclidian algorithm since thats what I have been learning.
Cheers.

2. Oct 3, 2012

### happysauce

Euclidean algorithm?

3. Oct 3, 2012

### epenguin

Well I think I got it, start with see why the chip bags have got to be an odd number and after that the chocolate bars have got to be...

Hopefully this leads to something interesting (because I can't say it starts that way )

4. Oct 3, 2012

### AlephZero

Notice that all three numbers are "nearly" multiples of 1.2.

You can use that fact to solve the problem without guessing, or chugging through Euclid's algorithm.

5. Oct 3, 2012

### HallsofIvy

Staff Emeritus
The first thing I would do it is multiply 12.30 = 1.20x + 2.50y by 10 to get 12x+ 25y= 123.

The only "Euclidean algorithm" needed is to note that 25- 2(12)= 1. Multiplying that equation by 123, 12(-246)+ 25(123)= 123. That is, one solution is x= -246 and y= 123. That is not the solution because -246 is not positive. But x= -246+ 25k and y= 123-12k is also a solution for ay integer, k: 12(-246+ 25k)+ 25(123- 12k)= 12(-246)+ (12)(25k)+ 25(123)- 25(12k)= 123 since the two terms in k cancel.

So you want to find an integer, k, such that x= -245+ 25k> 0 and y= 123- 12k> 0.

6. Oct 3, 2012

### the_Doctor111

Thanks heap guys! :!!)

7. Oct 3, 2012

### SammyS

Staff Emeritus
By the Way: Welcome to PF, Doctor111 !