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?

For the reasons articulated nicely by Ed, h must be 2. Now consider the left side, c*(c+2). This can be rewritten as (c+1)^2 - 1 since both are the same as c^2 + 2c.

(c+1)^2 - 1 = 120 + d or, isolating d,

(c+1)^2 - 121  = d

but the left side is the difference of squares, so

d = ((c+1) - 11)*((c+1)+11) = (c-10)*(c+12)

Yet d is prime. So one of these two terms must be equal to 1. Therefore, either c=11 or c=-13. Since c>0, c=11 is the only solution. d = (11-10)*(11+12) = 23

The only possible solution for (c,h,d) is (11,2,23)

 

Posted by Paul on 2008-10-20 21:01:56