Solving a Diophantine Equation with $4000 and Cows, Lambs & Piglets

[lokisapocalypse]

The question reads:

A farmer purchased 100 head of livestock for a total cost of $4000. Prices were as follow: calves, $120 each; lambs, $50 each; piglets, $25 each. If the farmer obtained at least one animal of each type, how many of each did he buy?

I obtained that -240 < t < -15.789 but this can't be right as there are way too many t's to check them all. Does anyone have a hint or some help?

 

[lokisapocalypse]

Never mind. Amazing how after looking at this problem for 4 hours, I post it here and 20 minutes later I see my mistake.

 

[robert Ihnot]

This one is sort of a brute force problem, but it can be simplified. We have X+Y+Z=100, where X=calves, Y=lambs, Z=piglets. Then for the money we have:

120X +50Y + 25Z=4000. This tells us that 5 divides X giving 5X'=X, and 2 divides Z giving 2Z' =Z. Thus form reduces to 12X' + Y+Z' = 80. Modifying the other equation gives 5X'+Y+2Z' =100. We can eliminate Y in one case, and going over the same equations eliminate Z'. Then we can look for more shortcuts or just use trial and error on the cases. Z', can not exceed 6 so there is no more than 6 cases to try.

And be careful about looking for ALL solutions!