The drunken jail guard
There is a circular jail with 100 cells numbered from 1 to 100 consectively. Each cell has an inmate and all the doors are locked to begin with. One night the jailor gets drunk and starts running around the jail in circles. In his first round he opens every door. In his second round he visits every 2nd door (2,4,6,...) and shuts the door. In the 3rd round he visits every 3rd door (3,6,9,...) and if the door is shut he opens it, if it is open he shuts it. This process of crazy drunken-ness continues for 100 rounds and finally the jailor gets exhausted and falls down. Let us assume the prisoners are all kind enough to wait until the crazy jailor stops and finally they decide to escape. Now how many prisoners find their doors open and can you also tell which cells' these prisoners belong to? People who post the answer should also post the logic behind their answers!
Level: Medium
Level: Medium
2 Comments:
the door will be open if the guard visits the room odd number of times.
tht would mean the num,ber should have odd number of factors.(including 1 and itself)
only possibilities are perfect squares.
so there will be a total of 10 prisoners who will escape.
and they are
1 4 9 16 25 36 49 64 81 and 100
Balaji TK,
Bingo, you got it perfectly right. This puzzle is SOLVED.
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