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 Sequence Group III (Posted on 2008-10-28)
Five positive integers A, B, C, D and E, with A < B < C < D < E, are such that:

(i) A, B and C (in this order) are in arithmetic sequence, and:

(ii) B, C and D (in his order) are in geometric sequence, and:

(iii) C, D and E (in this order) are in harmonic sequence.

Determine the minimum value of (E-A) such that there are precisely two quintuplets (A, B, C, D, E) that satisfy all the given conditions.

Note: Try to solve this problem analytically, although computer program/ spreadsheet solutions are welcome.

 See The Solution Submitted by K Sengupta No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
 re: computer solutions -- no proof | Comment 3 of 5 |
(In reply to computer solutions -- no proof by Charlie)

Using substitution I was able to derive the equation,
[E-A] = (4/A)*x2 + 4*x,
such that x is the common difference between A and B.

With this bit of information, and limiting my search to where
[E-A] = 64, I was able to search beyond A=200,000 to find if there were any subsequent sets. Yet, due to limitations, my search limited A to not much greater than 2,000,000, i.e., only a 10-fold increase in range.  No values for (A, x) other than (8, 8) and (36, 12) such that D also was an integer, i.e.,
x2 modulo (A + x) = 0, was found.
 Posted by Dej Mar on 2008-10-29 10:09:47

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