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 Sequence Group V (Posted on 2011-03-31)
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 harmonic sequence, and:

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

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

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

 No Solution Yet Submitted by K Sengupta Rating: 3.0000 (1 votes)

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 At last a winner Comment 4 of 4 |

After a digression into powers of 2, I now revert to my original line of thinking:

A    b   B   C   c    D    d     E   E-A
6   4   10  30  3   90   60 150  144
18   9   27  54  2   108  54 162 144
48 16   64  96 1.5 144  48 192  144

144=2^4*3^2 so I doubt there can be a smaller solution.Unfortunately, this is not the end of the story since on further research {A,b} {150,25} also produces a solution, for a total of 4, which is not precisely 3. The smallest number for which there are precisely 3 solutions, using the equality:

E=A(A+b)^2/(A-b)^2: E-A=(4A^2b)/(A-b)^2

is 192:

A    b   B    C    c       D     d     E    E-A
4    3    7    28  4      112   84  196  192
24   12   36   72  2      144   72  216  192
108   27 135  180 1.333 240   60  300  192

Edited on April 7, 2011, 4:07 am
 Posted by broll on 2011-04-06 12:26:51

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