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Geometric Sequence Settlement II (Posted on 2014-04-12) Difficulty: 3 of 5
The number 930 is divided by a positive integer d, giving a quotient of q and a remainder of r, with q < d, each of which is also a positive integer.

It is observed that the q, r and d are in geometric sequence.

Determine all possible triplets (q, r, d ) satisfying the given conditions and, prove that there are no others.

No Solution Yet Submitted by K Sengupta    
Rating: 3.0000 (1 votes)

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Solution Analytical Solution (spoiler) Comment 1 of 1
Let r = aq.  Then d =(a^2)q

930 = qd + r = a^2q^2 + aq = aq(aq + 1) = r(r+1)

solving qives r = 30

then qd = 900, and q can be any factor of 900 that is less than 30.  900 = 2^2 * 3*2 * 5*2 so the solutions are

(1,30,900)
(2,30,450)
(3,30,300)
(4,30,225)
(5,30,180)
(6,30,150)
(9,30,100)
(10,30,90)
(12,30,75)
(15,30,60)
(18,30,50)
(20,30,45)
(25,30,36)

Edited on April 12, 2014, 1:18 pm
  Posted by Steve Herman on 2014-04-12 11:34:57

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