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the basement => the archives => Logic Puzzles => Topic started by: Azron on January 23, 2004, 08:34:38 am

k. First things first with this puzzle is to look at what we know:
1) There are 16 coins that need to flipped
2) There must be a total of 18 flips (6x3)
3) Any one coin must be flipped an odd number of times
These 3 things tell us that one coin must be flipped 3 times, the remaining 15 to be flipped once (if more than one coin was flipped 3 times, there would be fewer flips than coins remaining).
Now, knowing every coin bar one can only be flipped once tells us that the coin that needs to be flipped 3 times must be on one of the main diagonals, and since the problem has rotation symmetry of order 4 we can concentrate on the leading diagonal alone.
If we number the coins thus:
01  02  03  04
05  06  07  08
09  10  11  12
13  14  15  16
Then taking the 01 as our coin to be flipped three times (being the first in the lead diagonal), we flip the following sets:
01, 02, 03
01, 06, 11
01, 05, 09
Which means coins 1, 2, 3, 5, 6, 9 and 11 are flipped. We then perform the moves:
07, 10, 13
04, 08, 12
14, 15, 16
Which flips the remaining coins as required.
Obviously this solution has slight variations (it can be rotated and performed in a different order, also, the way the last 9 coins to be flipped are arranged into 3coin groups can be changed slightly).

Nice puzzle, had me slightly baffled to begin with, but once I applied a bit of mathematically inspired logic it came out

I must say that I loved your use of math in solving this puzzle.