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?

Assuming that all prime numbers are odd, the sum c+h must be even, making the left side of the equation even.  But 120 + d must be odd.  Therefore, not all the primes can be odd, meaning that one must be '2'.  We know that 'c' cannot be '2' because that would make the left side even while the right side remained odd.  So that leaves 'h' and 'd'.

First consider d=2, or c*(c+d) = 122.  The problem here is that 122 has only two factors - 2 and 61.  Since the variables must be different, 'c' cannot equal 2.  Thus, h=2.

The left side can be rewritten using a new variable 'n' where:

n = c+1 and n = c+(2-1) = (c+h)-1

So c*(c+h) can be rewritten as (n+1)*(n-1)

or n²-1

So:  n²=121+d

Starting at 121, it is simply a matter of finding all the squares greater than 121 and subtracting 121 to see if a prime number results.

12²-121 = 23

Now, checking for n=12

c=11, h=2, d=23

Posted by hoodat on 2008-10-26 23:51:19