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
(EA) 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.
(In reply to
proofthanks to Dej Mar's formula by Charlie)
The equation, [EA] = (4/A)x^{2} + 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)16^{2 }+ 4(16)
[A/64*]64 = [A/64]*((4/A)16^{2} + 64)
A [ A] = 64 + A [ A]
0 = 64
Where x is equal or greather than 17, A has a negative value...
64 = (4/A)17^{2 }+ 4(17)
64 [ 68] = 1156/A + 68 [ 68]
[A/4]*4 = [A/4]*1156/A
A = 289
If we assign [EA] = 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 [EA] to equal 64. Charlie and I have confirmed that all values of [EA] less than 64 do not have two quintuplets, thus, in agreement with Charlie, 64 is the minimum value of [EA] that satisfy all conditions.
Edited on October 30, 2008, 8:05 am

Posted by Dej Mar
on 20081030 00:56:14 