1. What is the equation for this problem?
A farmer goes to a sale with exactly $100 and must buy exactly 100 animals,and he has to buy one of each. Pigs are$3, chickens are $.50, and cows are$10.  2.

3. Sounds like a linear diophantine problem to me

3 x + 1/2 y + 10 z = 100

x + y + z = 100

x, y, z all positive integers

Trivially there must be an even number of chickens, and then go from there...  4. Originally Posted by river_rat
3 x + 1/2 y + 10 z = 100

x + y + z = 100

x, y, z all positive integers

Trivially there must be an even number of chickens, and then go from there...
Actually, the problem says he must buy one of each (at least). So:

3(x+1) + 1/2(y+1) + 10(z+1) = 100

3 x + 1/2 y + 10 z = 86.5

x + y + z = 97

The solutions (x,y,z) will all be 1 less than the actual. So just add 1 to each.

However this fact might not actually change the solution.  5. x, y, z are postive integers covers the fact that at least one must be bought, (0 is not a positive integer) though just shifting everything up by one and using natural numbers (assuming 0 \in N that is ) wouldnt change the solution.  6. It is never stated that he has to spend exactly $100, only that he went with$100.  7. True, but he has to spend exactly $100  8. Originally Posted by river_rat x, y, z are postive integers covers the fact that at least one must be bought, (0 is not a positive integer) though just shifting everything up by one and using natural numbers (assuming 0 \in N that is ) wouldnt change the solution. Oh I missed your condition. I was just looking at the equations. But yes, I guess you are correct, too. The way I posted it is how I would have done it   9. please explain, if you can only buy one animal each. the highest amount of money you spend can only be$13.5  10. "He has to buy one of each" is satisfied as long as he buys at least one of each.  Bookmarks
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