A set of 47 disks are consecutively numbered 1 to 47 and placed in a row as follows: 1, 2, 3, 4, ... 45, 46, 47.

Rearrange the disks so for any two given disks A and B, the disk equal to their arithmetic mean doesn't lie between them. For example, Disk 4 cannot lie between Disk 1 and Disk 7 since the arithmetic mean of 1 and 7 is 4. However, since 7 is not equal to the arithmetic mean of 1 and 4, Disk 7 may lie between Disk 1 and Disk 4.

(In reply to solution by Daniel)

Daniel, I see some problems in your solution.  The disk with the arithemetic mean seems to lie between each pair of disks.  5 is the arithemetic mean of 1 and 9, 9 is the arithemetic mean of 5 and 13, etc.  The problem called for a solution where the arithemetic mean did not lie between any two disks.

I do believe I accomplished this solution. I first divided the disks numbered even and odd.  Then divided the each pile into two separate piles, placing every other disk into one and then the other pile.  Then, working with the odd piles, from each pile I laid them in the pattern 7-B-9-3-1-5-6-2-4-A-C-8, beginning with 1 and proceeding to C (or B), and then appending each row to the other.  The same was done to the even numbered disks, and those were then appended to the odd numbered disks.  The following sequence was then formed:

25, 41, 33, 9, 1, 17, 21, 5, 13, 37, 45, 29, 27, 43, 35, 11, 3, 19, 23, 7, 15, 39, 47, 31, 26, 42, 34, 10, 2, 18, 22, 6, 14, 38, 46, 30, 28, 44, 36, 12, 4, 20, 24, 8, 16, 40, 32

(Correction made for placement of disks 45, 46 & 47.)

Edited on July 4, 2006, 10:38 pm

Posted by Dej Mar on 2006-07-04 13:57:13