 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Quadruplet Query (Posted on 2014-08-14) 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.

 No Solution Yet Submitted by K Sengupta Rating: 3.0000 (1 votes) Comments: ( Back to comment list | You must be logged in to post comments.) Analytical Solution (spoiler) | Comment 1 of 2
```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 Please log in:

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