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Arrange The Disks (Posted on 2006-07-04) Difficulty: 4 of 5
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.

  Submitted by K Sengupta    
Rating: 3.6667 (3 votes)
Solution: (Hide)
FIRST METHOD (Given by Charlie):

Charlie's methodologies culminating in a comprehensive soluion to the problem under reference are given in location1; location2; and location3.

SECOND METHOD:

One of the arrangements is given by :

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

This is obtained in the following manner.

First, we arrange the disks from 1 to 47 left to right. This gives us one "group" of disks.

We now repeat the following for every group:

1. Pulling the first disk to the left to start a new group.

2. Adding every other disk to the new group in order.

3. We now have twice as many groups, each half the size of the first. The process is repeated until every group is of size 1 or 2.

Consequently, we obtain:

First group: 1-47

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

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

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

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

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

Consequently, the required arrangement is:

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

N.B.: Within each group it does not matter whether the odd or even ones go left or right as has been rightly pointed out by Charlie in this location.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
AcknowledgementK Sengupta2006-08-31 05:09:21
128 ways for 1 through 8Charlie2006-07-06 09:43:42
re: A sampling of the trillions of solutions.Charlie2006-07-06 09:09:24
SolutionA sampling of the trillions of solutions.Charlie2006-07-06 09:04:45
re: solution (if I understood correctly)brianjn2006-07-05 20:42:03
re: solution (if I understood correctly)Charlie2006-07-05 14:05:43
solution (if I understood correctly)atheron2006-07-05 13:57:46
Solutionre(2): solution (with solution/spoiler)Charlie2006-07-05 12:23:15
Left behindbrianjn2006-07-05 06:05:37
re(3): solution (with solution/spoiler)Dej Mar2006-07-05 00:49:16
re(2): solution (with solution/spoiler)Charlie2006-07-04 22:00:56
Solutionre: solution (with solution/spoiler)Dej Mar2006-07-04 13:57:13
re: solutionCharlie2006-07-04 13:43:37
solutionDaniel2006-07-04 11:51:22
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