Linear equation problem

Five hundred dollars were used to buy one hundred animals of three types. The burros cost 11 dollars each, the camels cost 9 dollars each, and the dogs cost 2 dollars each. How many animals of each kind were there?

This is a problem from a linear equation section of a number theory class I am in, and it have had no success in solving it.

Here's a start

as there are 100 animals

each animals cost and the total cost

I've successfully done that, but have been trying to figure out how to combine them or get a solution from them. I've mainly tried to combine them and solve for variables in an augmented matrix, but have had no luck so far.

noramlly you would need additional information to form a 3rd equation and solve them simultanuously.

Given the numbers for B,D & C are discrete whole numbers I would probably use trial to get a result.

You can explore this equation in 2 variables to find discrete solutions.

11b + 9c +2d = 500
b +c + d = 100

where a, b, c and all positive integers

the two equations above imply 9b + 7c = 300

Using Euclid's Extended Algorithm (I think that's what it's called) I found 5 solutions for (b,c): (31,3),(24,12),(17,21),(10,30), and (3,39).

So there are 5 solutions for (b,c,d): (31,3,66), (24,12,64),(21,17,62), (10,30,60), and (3,39,58).

Quote:

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Originally Posted by Random Variable

11b + 9c +2d = 500
b +c + d = 100

where a, b, c and all positive integers

the two equations above imply 9b + 7c = 300

Using Euclid's Extended Algorithm (I think that's what it's called) I found 5 solutions for (b,c): (31,3),(24,12),(17,21),(10,30), and (3,39).

So there are 5 solutions for (b,c,d): (31,3,66), (24,12,64),(21,17,62), (10,30,60), and (3,39,58).

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I know a variation of the Euclidean Algorithm, and I am curious as to what you did to get these 5 solutions. That does match with what my teacher told me, he said there would be five solutions.

To solve 9b+ 7c= 300:

7 divides into 9 once with remainder 2: 2= 9- 7

2 divides in 7 3 times with remainder 1: 1= 7- 3(2)

So 1= 7- 3(9- 7)= 4(7)- 3(9) (Of course: 28- 27= 1)

So, multiplying by 300, 1200(7)- 900(9)= 300. That means that b= -900, c= 1200 is a solution. Unfortunately, not a valid solution because a number of animals cannot be negative!

But b= -900+ 7k, c= 1200- 9k is also a solution for any integer k: 9b+ 7c= 9(-900+ 7k)+ 7(1200- 9k)= 9(-900)+ 7(1200)- 63k+ 63k= 300 as the "k" terms cancel.

In order that b= -900+ 7k be positive, k must be larger than 900/7: 129 is the smallest such number. In order that c= 1200- 9k be positive, k must be less than 1200/9: 133 is the largest such number. Take k= 129, 130, 131, 132, 133 to get the 5 valid answers.

Thanks for the help everyone :)