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| Geometric Sequence Settlement II (Posted on 2014-04-12) |
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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.
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No Solution Yet
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Submitted by K Sengupta
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Rating: 3.0000 (1 votes)
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Analytical Solution (spoiler)
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Comment 1 of 1
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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
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