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.

With a small sample I had noticed that it seemed probable to get a result by slitting the 47 disks into Odds and Evens, as previous respondents have done.

Now, in each of these sets I did note that, if Disk B was to be the mean of Disk A  and Disk C, and still be within that set, then the interval from Disk A to Disk C had to be 4*n [n being and integer].

I note Charlie has some 'triplets' with the odd numbers, in response to Daniel.  I am wondering why Charlie settled upon 115.

My triplet construct is [ x, x+2n, x+4n].

Within the limit of this problem I believe that a mean can be generated in 132 ways for the Odd set and 121 for the Even; in toto, 253!

Given such a weight of considerations, if Dej Mar truly has a correct solution, and it does appear good to me, I hope his solution did not to weigh into the interesting sojourns I took.

Posted by brianjn on 2006-07-05 06:05:37