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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.

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

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solution (if I understood correctly) | Comment 8 of 14 |

it was pretty easy to see that the disks could be placed correctly by placing odds and evens in right spots. I started with 1,gap,2,gap,3... all the way to 24 and then filling those gaps starting from the beginning with 25 and so on...

The order was like this:

1,24,2,25,3,26,4,27,5...21,44,22,45,23,46,24,47

Correct me if there's something wrong.. The solution felt so easy that I thought I understood something wrong..


  Posted by atheron on 2006-07-05 13:57:46
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