Jiaqi Guo's home
Hmm... unfortunately the parity of the initial conditions the switches can be in is random. This solution is very similar to the idea you had before - flip the first switch and change the parity on the first time you enter the room, flip the second switch and leave the parity constant otherwise.
Thursday, January 24, 2008
Jiaqi Guo's home
Jiaqi Guo's home
Because there are two bits, the switches can represent the numbers 0,1,2,3 in binary, and it is possible for each person to flip a switch and either change the parity of the number, or leave it unchanged. For instance one flip will take 0 to 2, or 0 to 1; 1 to 0 or 1 to 3; 2 to 0 or 2 to 3, etc. Try working out a protocol where each prisoner knows what to do when he finds a given parity for the first, second, third,.... time he enters the room.
Because there are two bits, the switches can represent the numbers 0,1,2,3 in binary, and it is possible for each person to flip a switch and either change the parity of the number, or leave it unchanged. For instance one flip will take 0 to 2, or 0 to 1; 1 to 0 or 1 to 3; 2 to 0 or 2 to 3, etc. Try working out a protocol where each prisoner knows what to do when he finds a given parity for the first, second, third,.... time he enters the room.
Sunday, October 14, 2007
Divisibility
What is the remainder if 1 + 2^2 + 3^3 + ... + 10^10 is divided by 2,3,4,5,6,7,8 and 9?
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