All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Just Math
Sequence Group III (Posted on 2008-10-28) Difficulty: 3 of 5
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

Please log in:
Remember me:
Sign up! | Forgot password

Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (2)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Copyright © 2002 - 2021 by Animus Pactum Consulting. All rights reserved. Privacy Information