100 animals 100 dollars? easy? I thought so.

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The discussion revolves around a mathematical problem involving purchasing 100 animals for exactly $100, with specific costs for cows, pigs, and chickens. Participants explore various equations and constraints to find combinations of animals that satisfy the conditions. The challenge lies in ensuring that the total number of animals and their costs align correctly, particularly with the pricing of chickens at $0.50. Several attempts are made to solve the problem through trial and error, leading to different potential solutions. Ultimately, the complexity increases due to the need for integer values and the specific pricing structure.
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I have to buy 100 animals, I have to use 100 dollars, I have to get at least 1 of each.

Cow(X) = 10 bucks
Pig(Y) = 1 buck
Chicken(Z) = .5 bucks, 50 cents, 2 quarters

This looked really easy at first but I kept getting stuck, how do I get through this?

Restraints I came up with.

x+y+z=100
10x+1y+.5z=100
X is greater than or equal to 1 and is less than or equal to 10
Y is greater than or equal to 1
Z is greater than or equal to 1
 
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does the total cost have to be $100 also??
 
i have the answer!

hint:

x (cows) can be reasonably guessed.

then you go from there

(this is how i did it. trial and error.lol)

only problem is, if you have to show work in terms of math, I am not sure.

i just tried out various values. took me a couple minutes.
 
Your system is fine. Solve it to get

x - \frac{z}{18} = 0, \; y + \frac{19}{18}z = 100

so you can choose z=18 \Longrightarrow x = 1, \; y = 81.
 
Data: so you can choose z=18,x=1,y=81
But, that fails to equal $100, since we have 10x1+1x81+.05x18 =$91.90.

Frankly, the number of chicken must be a multiple of 20, since the other monetary values are in dollars. Then while we are at it, the cost of the pigs and the chickens must be a multiple of $10.
 
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nope ... the number of chickens is not a multiple of 20
 
If the total cost is to be $100 how are we to get rid of the odd nickels if the chickens are not a multiple of 20? Are you planning to add a tax?
 
its not nickels, but $0.50
 
oops, i used $0.5 for z~

There's no positive integer z that also makes x and y positive integers that satisfies the system if it's really $0.005.
 
  • #10
at least for my answer and what he wrote above (.5) which i took as 50 cents
 
  • #11
i had a different answer

4 cows = $40
24 pigs = $24
72 chickens = $36
===============
100 animals at $100
 
  • #12
It's even worse than nickels. The way that he stated it it was half-pennies.
 
  • #13
Yes, there are several possible answers if a chicken costs $0.5.
 
  • #14
I get:

x,y,z
1, 81, 18
2, 62, 36
3, 43, 54
4, 24, 72
5, 5, 90
 
  • #15
woops, lol ahaha. I feel retarded. chickens are 50 cents, or 2 quarters. Wow sorry about that.

note- made a few changes to my first post to answer some of your questions
 
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