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Cows, horses and dogs (Posted on 2008-10-20) Difficulty: 2 of 5
I have cows, horses and dogs, a different prime number of each. If I multiply the number of cows (c) 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 of each do I have?

See The Solution Submitted by pcbouhid    
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Solution computer solution | Comment 2 of 12 |

As c^2 + ch - (120 + d) = 0, the quadratic formula can find c for a given h and d.

c = (-h + sqrt(h^2 + 4(120 + d)))/2

Then c can be tested for primality.

The UBASIC program evaluates this allowing h and d to be up to the 10,000th prime number:

 10   for Hct=1 to 10000
 20    for Dct=1 to 10000
 30      H=prm(Hct)
 40      D=prm(Dct)
 50      Disc=H*H+4*(D+120)
 60      Sr=int(sqrt(Disc)+0.5)
 70      if Sr*Sr=Disc and (Sr-h)@2=0 then
 80        :C=int((Sr-H)/2)
 90        :if prmdiv(C)=C then
100           :print C,H,D
110    next
120   next

finds two solutions, one of which is spurious as it has the same number of dogs as cows, contrary to the problem statement. The spurious one is 2 cows, 59 horses and 2 dogs.

The intended solution is 11 cows, 2 horses and 23 dogs.

Note: In UBASIC, prm(x) returns the xth prime number, and prmdiv(x) returns the smallest prime divisor of x. The prmdiv function is applicable only to numbers in integer form--not to real numbers even if they happen to have an integral value, hence line 80, converting a real that happens to be an integer to integer format.

  Posted by Charlie on 2008-10-20 12:22:28
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