Find four positive integers p,q,r and s such that:

• p*q*r*s= 8!, and:
• p*q+p+q = 524, and:
• q*r+q+r = 146, and:
• r*s+r+s = 104

Prove that there are no others.

a) p*q*r*s = 8!, so the only prime factors in these numbers are 7 and less. 

b) Subtracting equation 4 from 3 and rearranging terms gives
   (r+1)(q-s) = 42,
   so (r+1) must be 2,3,6,7,14,21 or 42
   This makes r = 1,2,5,6,13,20 or 41. 
   We can rule out 13 and 41 (prime factors are too large),
   so r = 1,2,5,6 or 20
   
c) From equation 3, q = (146-r)/(1+r)
   The possibilities are
   r  q
   -- --
   1  72.5
   2  48
   5  23.5
   6  20
   20  6
   
   q must be integral, so q must be 48 or 20 or 6
   
 d) from equation 2, p = (524-q)/(1+q)
   The possibilities are
   r  q  p
   -- -- --
   2  48 9.714286..
   6  20 24
   20  6 74
   
   of these 3, we can rule out the fractional p and the multiple of 37, 
   so the only solution is
   
   r  q  p  s
   -- -- -- --
   6  20 24 14
   
   A quick check confirms that their product = 40320 = 8!

Edited on August 14, 2014, 1:04 pm

Posted by Steve Herman on 2014-08-14 13:02:46