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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?

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Solution Puzzle Solution Comment 12 of 12 |
(In reply to Answer by K Sengupta)

It is given that:

c(c + h) = 120 + d .......(i)

If d is even, then it follows that d must be equal to 2, since d is the only even prime.

So, c(c+h) = 122. Since no prime factor other than 2 divides 122, it follows that: c = 2, which is a contradiction since by rhe given conditions c and d must be different.

Therefore, it follows that d is odd. If both c and h are odd, then in that situation, (c+h) is even so that the rhs of (i) is even, which diretly contradicts the assumption that d is odd.
Similarly, both c and h cannot be even, since in that situation c=h=2, a contradiction.

Consequently, precisely one of c and h is even, while the other is odd.

If c is even, then that would give and even number in the rhs of (i). this is a contradiction.

Thus, h must be even, so that h=2, and therefore:

c(c+2) = 120 + d
-> (c+12)(c-10) = d ......(ii)

If c>=11, the rhs of (ii) would be composite. This is a contradiction. Also, c<=10, would force a nonpositive value in the rhs of (i).
Accordingly, c-10 =1, giving: c=11, so that: d=23
Therefore, (c,h,d) = (11,2,23) is the only possible solution.

Consequently, the respective numbers of cows, horses and dogs are 11, 2 and 23.

  Posted by K Sengupta on 2008-11-21 05:05:24
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