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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: proof--thanks to Dej Mar's formula Comment 5 of 5 |
(In reply to proof--thanks to Dej Mar's formula by Charlie)

The equation, [E-A] = (4/A)x2 + 4x, can show that A must be much less than even 16,384 for 64 to be the solution.

x must be < 16.
If the common difference is assigned as equal to 16, we get an invalid equality, and A is undefined...
64 = (4/A)162 + 4(16)
[A/64*]64 = [A/64]*((4/A)162 + 64)
A [- A] = 64 + A [- A]
0 = 64
Where x is equal or greather than 17, A has a negative value...
64  = (4/A)172 + 4(17)
64 [- 68] = 1156/A + 68 [- 68]
[-A/4]*-4 = [-A/4]*1156/A
A = -289

If we assign [E-A] = 64 and x to the next smaller value, 15...
64 = (4/A)(15)2 + 4(15)
64 [- 60] = 900/A + 60 [- 60]
[A/4*]4 = [A/4*]900/A
A = 900/4 = 225

Thus, A can not be greater than 225 in order for [E-A] to equal 64. Charlie and I have confirmed that all values of [E-A] less than 64 do not have two quintuplets, thus, in agreement with Charlie, 64 is the minimum value of [E-A] that satisfy all conditions.

Edited on October 30, 2008, 8:05 am
 Posted by Dej Mar on 2008-10-30 00:56:14

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