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Quadruplet Query (Posted on 2014-08-14) Difficulty: 2 of 5
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.)
Negative solutions Comment 2 of 2 |
By the way, if negative integers had been allowed, there would still be only one solution

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 if r is negative then (r+1) must be -1,-2,-3,-6,-7,-14,-21 or -42
   This makes r = -2,-3,-4,-7,-8,-15,-22 or -43. 
   We can rule out -22 and -43 (prime factors are too large),
   so r = -2,-3,-4,-7,-8, or -15
   
c) From equation 3, q = (146-r)/(1+r)
   The possibilities are
   r   q
   --  --
   -2 -148  <== multiple of 37 
   -3 -74.5 <== not an integer
   -4 -50   <== multiple of 25
   -7 -25.5 <== not an integer
   -8 -22   <== multiple of 11
   -15 -11.5 <== not an integer
   
so, r must be positive, and the only solution is the one previously posted

  Posted by Steve Herman on 2014-08-14 14:32:17
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