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**