Tough math problem my uncle approached me with...

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xXAceFireXx

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You have $1000 and you have to go out and buy chickens, donkeys, and cows.

Chickens cost $10
Donkeys cost $15
Cows cost $20

You have to use up the entire $1000 exactly and you have to buy a total of 75 animals. You also have to buy at least one of each. What combination of animals do you buy?

I've been trying to figure this one out for a couple days now, but im not gettin anywhere.
 
\(\displaystyle \L\\10x+15y+20z=1000\)...[1]

\(\displaystyle \L\\x+y+z=75\)...[2]

You have 3 variables but only 2 equations.

Solve [2] for y and sub into [1] and get:

\(\displaystyle \L\\-5x+5z+1125=1000\)

Which reduces to \(\displaystyle x-z=25\)

Try some values.

30 chickens, 5 cows and 40 donkeys.

That's 75 critters and

30*10+5*20+40*15=1000.
 
wow very nice...i figured guess and check would work after a little simplification, but im still a little unsatisfied.

is their any way this can be solved completely algebraically? or do you have to use guess and check because of the nature of this problem?
 
You have $1000 and you have to go out and buy chickens, donkeys, and cows.

Chickens cost $10
Donkeys cost $15
Cows cost $20

You have to use up the entire $1000 exactly and you have to buy a total of 75 animals. You also have to buy at least one of each. What combination of animals do you buy?
Click to expand...

This problem has multiple solutions. For example:
49 chickens, 2 donkeys, 24 cows
48 chickens, 4 donkeys, 23 cows

The key here is that the number of donkeys must be even. Note also that 1 cow + 1 chicken = 2 donkeys (in value/cost). Therefore a pattern emerges: decrease the number of chickens and cows by 1 each and increase the number of donkeys by 2.
 
As wjm pointed out, you have 3 variables and 2 equations. Therefore, there are multiple solutions. I just chose a nice, neat one.
 
Bet your uncle 10 bucks there's 24 solutions;
cows:donkeys:chickens:

1 : 48 : 26
2 : 46 : 27
3 : 44 : 28
...
23: 4 : 48
24: 2 : 49
 
You have $1000 and you have to go out and buy chickens, donkeys, and cows.

Chickens cost $10
Donkeys cost $15
Cows cost $20

You have to use up the entire $1000 exactly and you have to buy a total of 75 animals. You also have to buy at least one of each. What combination of animals do you buy?

1-- c + D + C = 75
2--10c + 15D += 20C = 1000
3--10c + 15D + 20C = 750
4--Subtracting, D + 2C = 50
5--D must be an integer k
6--Therefore, 5k + 2C = 50 makingC = (50 - k)/2
7--k...1...2...3...4...5...6...7...8...9.......etc.
...D...1...2...3...4...5...6...7...8...9.......etc.
...C...-..24...-..23...-..22...-..21...-.......etc.
...c......49...-..48...-..47...-..46...-.......etc.

for a total of 24 solutions.
 
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