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Cows, horses and dogs (Posted on 2008-10-20) |
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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?
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Submitted by pcbouhid
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Rating: 4.0000 (1 votes)
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Solution:
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The equation c*(c + h) = 120 + d reveals that d cannot equal 2, the only even prime number, since C would then also have to be 2, which is not a different number.
So d must be odd, from which either c or h must equal 2, since the sum of the two would otherwise be even.
Clearly, h must then must be 2, since c equals 2 would made the product even. Rewriting the equation, we get:
c2 + 2c - 120 = d (c + 12)(c - 10) = d
(c - 10) must equal unity in order for d to be a prime. Then:
c = 11, h = 2 and d = 23 is the unique solution. |
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