Solving x^{20} + y^{10} = 2000 - Bob

[Bob19]

Hi

I'm told that the following can be viewed as a an euler problem:

If a farmer goes to marked and wants to buy chickens and cows for 2000 dollars.

One chicken costs 20 dollars and one cow 10 dollars. How many chickens and cows does that farmer buy ?

Is it $$x^{20} + y^{10} \equiv 2000$$ ?

if yes how do I solve it ?

/Bob

 

[TD]

If a chicken costs 20 and you represent the number of chickens by x, the cost would be 20x instead of x^20, no? Same for the cows, I don't see why you introduced powers here.

Edit: bit too late

 

[Bob19]

Okay it is then
$$20x + 10y = 2000$$ ?

If a chicken costs 20 and you represent the number of chickens by x, the cost would be 20x instead of x^20, no? Same for the cows, I don't see why you introduced powers here.
Edit: bit too late

 

[TD]

That seems more logical, yes

 

[JaysonL]

If a farmer goes to marked and wants to buy chickens and cows for 2000 dollars.

One chicken costs 20 dollars and one cow 10 dollars. How many chickens and cows does that farmer buy ?

Since there hasn't been a solution, I'll post one for those who are interested.

Let x be the number of chickens which the farmer buys.
Let y be the number of cows which the farmer buys.

Therefore,
20x + 10y = 2000
2x + y = 200

2x= 200-y ... (1)
x= (200-y)/2 = 100 - y/2

Since x is an integer and 100, is an integer.
Therefore y/2 must be an integer.

Let y= 2a, where a is any integer.

Substitute y=2a into (1):
2x= 200 - 2a
x = 100-a

General solutions are:
x=100-a ... (2)
y=2a ... (3)
-----------------------------------------------
For positive integer solutions:

Let x = 100-a > 0, a< 100
Let y = 2a > 0, a>0

Therefore a can have any integer value between 1 - 99 inclusive.
-----------------------------------------------------
To find 1 solution, substitute one value of a into (2) and (3): (For example, use a =1 )
x = 100 - a = 100- 1 = 99
y = 2a = 2(1) = 2

One solution for this question would be:
The farmer buys 99 chickens and 2 cows.