A set of 47 counters is consecutively numbered from 1 to 47 and placed in a row as follows: 1, 2, 3, 4, ... 45, 46, 47.
Rearrange the counters so that, for any two given counters A and B, the counter equal to their arithmetic mean doesn't lie between them.
For example, Counter 4 cannot lie between Counter 1 and Counter 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, Counter 7 may lie between Counter 1 and Counter 4.
Solution to the Problem:
There are numerous solutions to this problem, but in each case, the 24 odd numbers must be grouped together and the 23 even numbers must be grouped together.
Here are some arrangements that work:
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
13, 45, 29, 21, 5, 9, 1, 37, 41, 25, 33, 17, 31, 15, 23, 7, 3, 47, 39, 35, 43, 19, 11, 27, 46, 14, 30, 22, 6, 10, 2, 38, 42, 26, 34, 18, 32, 16, 24, 8, 4, 40, 36, 20, 12, 44, 28
1, 33, 17, 9, 41, 25, 5 ,37, 21, 13, 45, 29, 3, 35, 19, 11, 43, 27, 7, 39, 23, 15, 47, 31, 46, 14, 30, 22, 6, 10, 2, 38, 42, 26, 34, 18, 32, 16, 24, 8, 4, 40, 36, 20, 12, 44, 28
9, 1, 41, 25, 33, 17, 21, 5, 13, 37, 45, 29, 43, 11, 3, 27, 35, 19, 31, 47, 15, 23, 7, 39, 10, 2, 42, 26, 34, 18, 22, 6, 14, 38, 46, 30, 44, 12, 4, 28, 36, 20, 32, 16, 24, 8, 40
2, 34, 18, 10, 42, 26, 6, 38, 22, 14, 46, 30, 4, 36, 20, 12, 44, 28, 8, 40, 24, 32, 16, 1, 33, 17, 9, 41, 25, 5, 37, 21, 13, 45, 29, 3, 35, 19, 11, 43, 27, 7, 39, 23, 31, 15, 47
K. Sengupta gives the following explanation for the arrangement
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
First, we arrange the counters from 1 to 47, left to right. This gives us one "group" of counters.
We now repeat the following for every group:
1. Pulling the first counter to the left to start a new group.
2. Adding every other counter to the new group in order.
3. We now have twice as many groups, each about 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.
Many thanks to Colin Bowey for sending me the Excel Spreadsheet to check the answers!!!
Click here to download Colin's spreadsheet
Correctly solved by:
| 1. Colin (Yowie) Bowey | Beechworth, Victoria, Australia |
| 2. Ritwik Chaudhuri | Santiniketan, West Bengal, India |
| 3. Kelly Stubblefield | Mobile, Alabama |