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A pet and prime problem

  1. Feb 17, 2009 #1
    Bert has some cows, horses and dogs, a different prime number of each.

    If the number of cows (c) is multiplied by the total of cows and horses (c+h), the product is 120 more than the number of dogs (d), that is:
    c*(c+h) = 120 + d.

    How many cows, horses and dogs does Bert have?
     
    Last edited: Feb 17, 2009
  2. jcsd
  3. Feb 17, 2009 #2

    HallsofIvy

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    I can't think of any "formula" for solving something like this but "guess and check" is an old, respected method for solving problems. 120 is about 11 squared so start by trying numbers around that. If c= 11, h= 3 c+h= 13, c(c+h)= 11(13)= 143= 120+ 23 and 23 is prime! c= 11, h= 3, d= 23 works.
     
  4. Feb 17, 2009 #3
    Of course, h=even and c=even is impossible.

    If h=2 , c=odd
    c**2 + 2c = 120 + d
    (c+1)**2 = 11**2 + d , so d is not prime.

    If h=odd , c=odd, then d=2 , and c*(c+h) = 2*61 , which is impossible.

    If c=2 and h=odd, then d=2. Then, c=2 and h=59.

    :smile:

    Edited:

    Of course, I should have noted "c" and "d" cannot be 2 ( Remember: "a different prime number").
    And then, I should have seen d=(c+12)(c-10) is prime if c=11, which implies d=23, and h=2.
     
    Last edited: Feb 18, 2009
  5. Feb 18, 2009 #4
    This is incorrect. It yields,
    d = (c-10)(c+12)
    which is prime if c=11. So h=2, c=11, d =23 is a solution (I assume this is the one HallsofIvy meant).
     
  6. Feb 18, 2009 #5

    HallsofIvy

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    I have edited my post so I can pretend I didn't make that mistake!
     
  7. Feb 18, 2009 #6

    Borek

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    What about 2 cows, 59 horses and 2 dogs?
     
  8. Feb 18, 2009 #7

    Borek

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    Interestingly, no more solutions for primes < 100000.
     
  9. Feb 18, 2009 #8

    uart

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    Borek, the question says that each of the three primes is different.

     
  10. Feb 18, 2009 #9

    Borek

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    Ah OK, somehow missed this condition.
     
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