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Diophantine Equation

  1. Feb 16, 2005 #1
    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 any one have a hint or some help?
  2. jcsd
  3. Feb 16, 2005 #2
    Never mind. Amazing how after looking at this problem for 4 hours, I post it here and 20 minutes later I see my mistake.
  4. Feb 21, 2005 #3
    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!
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