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 re: solution (with solution/spoiler) by Dej Mar)

Upon seeing Dej Mar's method of arranging the numbers, I thought of representing the numbers as binary. The numbers, shown in the order of Dej Mar's latest modification, with their binary equivalents are:

25 011001
41 101001
33 100001
 9 001001
 1 000001
17 010001
21 010101
 5 000101
13 001101
37 100101
45 101101
29 011101
27 011011
43 101011
35 100011
11 001011
 3 000011
19 010011
23 010111
 7 000111
15 001111
39 100111
47 101111
31 011111
26 011010
42 101010
34 100010
10 001010
 2 000010
18 010010
22 010110
 6 000110
14 001110
38 100110
46 101110
30 011110
28 011100
44 101100
36 100100
12 001100
 4 000100
20 010100
24 011000
 8 001000
16 010000
40 101000
32 100000

The initial odd/even split produces the separation of 1's from 0's at the right end. Within each group, the 1's and 0's from the next-to-rightmost bit are again separated, and similarly for the third from right. I notice that starting with the fourth from the right he departs from this, placing 33 before the 9, the former then being a zero among 1's in the fourth position from the right.  I wonder if this is made possible by the numbers not going all the way to 63 (111111).

Here is a set that works from 0 through 63, shown with the binary representation in reverse (units position first and 32's position last).

 0 000000
32 000001
16 000010
48 000011
 8 000100
40 000101
24 000110
56 000111
 4 001000
36 001001
20 001010
52 001011
12 001100
44 001101
28 001110
60 001111
 2 010000
34 010001
18 010010
50 010011
10 010100
42 010101
26 010110
58 010111
 6 011000
38 011001
22 011010
54 011011
14 011100
46 011101
30 011110
62 011111
 1 100000
33 100001
17 100010
49 100011
 9 100100
41 100101
25 100110
57 100111
 5 101000
37 101001
21 101010
53 101011
13 101100
45 101101
29 101110
61 101111
 3 110000
35 110001
19 110010
51 110011
11 110100
43 110101
27 110110
59 110111
 7 111000
39 111001
23 111010
55 111011
15 111100
47 111101
31 111110
63 111111

Note that in this instance if the 33 and 9 are reversed, the 9 would fall between the 1 and the 17, the 17 between the 1 and the 33 and also between the 9 and the 25, and the 33 between the 9 and the 57.

Of course if you only need 1 through 47, then just leave out the other numbers.

Other arrangements are possible, as at each stage it does not matter if the zeros come first or the ones.  Consistency was maintained here for ease of sorting the reversed binary numbers.

Posted by Charlie on 2006-07-05 12:23:15