Solution to Puzzle #80: The Bored Guards

While the puzzle did not require any special mathematical knowledge, it was a relatively difficult puzzle and I did not get as many correct answers. The only correct asnwer was from Girish Tutakne.

Here is a simple explanation – Number the towers from 1 to 6. Initially we have one guard on every tower. If we add the tower # for each guard, that would initially be 21 (sum of 1 to 6), which is an odd parity. When any one guard moves, they move from either an odd to an even tower or vice versa, in either case the parity of the sum changes. However, since two guards move every 15 minutes, the parity remains the same as it was in the beginning.

For all the guards to be on the same tower, the parity would be even (6 multiplied by any of the tower #). However, we begin with an odd parity and continue to be odd parity after every move, and hence can never achieve the situation of all the guards in the same location.

Hope you enjoyed the puzzle!

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