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 Arrange The Disks (Posted on 2006-07-04)
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.

 See The Solution Submitted by K Sengupta Rating: 3.6667 (3 votes)

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 128 ways for 1 through 8 | Comment 13 of 14 |

7, 3, 5, 1, 8, 4, 6, 2
6, 2, 8, 4, 7, 3, 5, 1
5, 1, 7, 3, 8, 4, 6, 2
8, 4, 6, 2, 5, 1, 7, 3
7, 3, 5, 1, 6, 2, 8, 4
6, 2, 8, 4, 5, 1, 7, 3
5, 1, 7, 3, 6, 2, 8, 4
4, 8, 6, 2, 7, 3, 5, 1
3, 7, 5, 1, 8, 4, 6, 2
2, 6, 8, 4, 7, 3, 5, 1
1, 5, 7, 3, 8, 4, 6, 2
4, 8, 6, 2, 5, 1, 7, 3
3, 7, 5, 1, 6, 2, 8, 4
2, 6, 8, 4, 5, 1, 7, 3
1, 5, 7, 3, 6, 2, 8, 4
8, 4, 2, 6, 7, 3, 5, 1
7, 3, 1, 5, 8, 4, 6, 2
6, 2, 4, 8, 7, 3, 5, 1
5, 1, 3, 7, 8, 4, 6, 2
8, 4, 2, 6, 5, 1, 7, 3
7, 3, 1, 5, 6, 2, 8, 4
6, 2, 4, 8, 5, 1, 7, 3
5, 1, 3, 7, 6, 2, 8, 4
4, 8, 2, 6, 7, 3, 5, 1
3, 7, 1, 5, 8, 4, 6, 2
2, 6, 4, 8, 7, 3, 5, 1
1, 5, 3, 7, 8, 4, 6, 2
4, 8, 2, 6, 5, 1, 7, 3
3, 7, 1, 5, 6, 2, 8, 4
2, 6, 4, 8, 5, 1, 7, 3
1, 5, 3, 7, 6, 2, 8, 4
8, 4, 6, 2, 3, 7, 5, 1
7, 3, 5, 1, 4, 8, 6, 2
6, 2, 8, 4, 3, 7, 5, 1
5, 1, 7, 3, 4, 8, 6, 2
8, 4, 6, 2, 1, 5, 7, 3
7, 3, 5, 1, 2, 6, 8, 4
6, 2, 8, 4, 1, 5, 7, 3
5, 1, 7, 3, 2, 6, 8, 4
4, 8, 6, 2, 3, 7, 5, 1
3, 7, 5, 1, 4, 8, 6, 2
2, 6, 8, 4, 3, 7, 5, 1
1, 5, 7, 3, 4, 8, 6, 2
4, 8, 6, 2, 1, 5, 7, 3
3, 7, 5, 1, 2, 6, 8, 4
2, 6, 8, 4, 1, 5, 7, 3
1, 5, 7, 3, 2, 6, 8, 4
8, 4, 2, 6, 3, 7, 5, 1
7, 3, 1, 5, 4, 8, 6, 2
6, 2, 4, 8, 3, 7, 5, 1
5, 1, 3, 7, 4, 8, 6, 2
8, 4, 2, 6, 1, 5, 7, 3
7, 3, 1, 5, 2, 6, 8, 4
6, 2, 4, 8, 1, 5, 7, 3
5, 1, 3, 7, 2, 6, 8, 4
4, 8, 2, 6, 3, 7, 5, 1
3, 7, 1, 5, 4, 8, 6, 2
2, 6, 4, 8, 3, 7, 5, 1
1, 5, 3, 7, 4, 8, 6, 2
4, 8, 2, 6, 1, 5, 7, 3
3, 7, 1, 5, 2, 6, 8, 4
2, 6, 4, 8, 1, 5, 7, 3
1, 5, 3, 7, 2, 6, 8, 4
8, 4, 6, 2, 7, 3, 1, 5
7, 3, 5, 1, 8, 4, 2, 6
6, 2, 8, 4, 7, 3, 1, 5
5, 1, 7, 3, 8, 4, 2, 6
8, 4, 6, 2, 5, 1, 3, 7
7, 3, 5, 1, 6, 2, 4, 8
6, 2, 8, 4, 5, 1, 3, 7
5, 1, 7, 3, 6, 2, 4, 8
4, 8, 6, 2, 7, 3, 1, 5
3, 7, 5, 1, 8, 4, 2, 6
2, 6, 8, 4, 7, 3, 1, 5
1, 5, 7, 3, 8, 4, 2, 6
4, 8, 6, 2, 5, 1, 3, 7
3, 7, 5, 1, 6, 2, 4, 8
2, 6, 8, 4, 5, 1, 3, 7
1, 5, 7, 3, 6, 2, 4, 8
8, 4, 2, 6, 7, 3, 1, 5
7, 3, 1, 5, 8, 4, 2, 6
6, 2, 4, 8, 7, 3, 1, 5
5, 1, 3, 7, 8, 4, 2, 6
8, 4, 2, 6, 5, 1, 3, 7
7, 3, 1, 5, 6, 2, 4, 8
6, 2, 4, 8, 5, 1, 3, 7
5, 1, 3, 7, 6, 2, 4, 8
4, 8, 2, 6, 7, 3, 1, 5
3, 7, 1, 5, 8, 4, 2, 6
2, 6, 4, 8, 7, 3, 1, 5
1, 5, 3, 7, 8, 4, 2, 6
4, 8, 2, 6, 5, 1, 3, 7
3, 7, 1, 5, 6, 2, 4, 8
2, 6, 4, 8, 5, 1, 3, 7
1, 5, 3, 7, 6, 2, 4, 8
8, 4, 6, 2, 3, 7, 1, 5
7, 3, 5, 1, 4, 8, 2, 6
6, 2, 8, 4, 3, 7, 1, 5
5, 1, 7, 3, 4, 8, 2, 6
8, 4, 6, 2, 1, 5, 3, 7
7, 3, 5, 1, 2, 6, 4, 8
6, 2, 8, 4, 1, 5, 3, 7
5, 1, 7, 3, 2, 6, 4, 8
4, 8, 6, 2, 3, 7, 1, 5
3, 7, 5, 1, 4, 8, 2, 6
2, 6, 8, 4, 3, 7, 1, 5
1, 5, 7, 3, 4, 8, 2, 6
4, 8, 6, 2, 1, 5, 3, 7
3, 7, 5, 1, 2, 6, 4, 8
2, 6, 8, 4, 1, 5, 3, 7
1, 5, 7, 3, 2, 6, 4, 8
8, 4, 2, 6, 3, 7, 1, 5
7, 3, 1, 5, 4, 8, 2, 6
6, 2, 4, 8, 3, 7, 1, 5
5, 1, 3, 7, 4, 8, 2, 6
8, 4, 2, 6, 1, 5, 3, 7
7, 3, 1, 5, 2, 6, 4, 8
6, 2, 4, 8, 1, 5, 3, 7
5, 1, 3, 7, 2, 6, 4, 8
4, 8, 2, 6, 3, 7, 1, 5
3, 7, 1, 5, 4, 8, 2, 6
2, 6, 4, 8, 3, 7, 1, 5
1, 5, 3, 7, 4, 8, 2, 6
4, 8, 2, 6, 1, 5, 3, 7
3, 7, 1, 5, 2, 6, 4, 8
2, 6, 4, 8, 1, 5, 3, 7
1, 5, 3, 7, 2, 6, 4, 8
8, 4, 6, 2, 7, 3, 5, 1

This uses 3-bit numbers to represent 0-7, and then prints out as 1-8 by adding 1.  There are 7 choice points to choose whether to place 0 bits or 1 bits first, and all combinations of those were used for the 2^7=128 orders.

 Posted by Charlie on 2006-07-06 09:43:42

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