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_ P Clue
NOTE THAT IT'S THE AGENT WITH THE GREEN BADGE AND NOT AGENT GREEN TALKING TO AGENT RED. AS THE AGENT WITH THE GREEN BADGE HAS SPOKEN TO AGENT RED, WE KNOW THAT AGENT RED DOESN'T HAVE A GREEN BADGE. WE ALREADY KNOW HE DOESN'T HAVE A RED BADGE. THEREFORE HE HAS A YELLOW BADGE. THE AGENT WITH THE GREEN BADGE CANNOT BE AGENT RED, NOR CAN HE BE AGENT GREEN, THEREFORE HE IS AGENT YELLOW. SO AGENT YELLOW HAS A GREEN BADGE, AND AGENT RED HAS A YELLOW BADGE, MEANING AGENT GREEN MUST HAVE A RED BADGE WHEN THE FIRST SERVANT COMES IN, THE KING SHOULD WRITE DOWN HIS NUMBER. FOR EACH OTHER SERVANT THAT REPORTS IN, THE KING SHOULD ADD THAT SERVANT'S NUMBER TO THE CURRENT NUMBER WRITTEN ON THE PAPER, AND THEN WRITE THIS NEW NUMBER ON THE PAPER. LET X BE THE NUMBER OF THE MISSING SERVANT AND Y BE THE NUMBER THAT THE KING HAS WRITTEN. ONCE THE FINAL SERVANT HAS REPORTED IN, THE NUMBER ON THE PAPER SHOULD EQUAL: Y = (1 + 2 + 3 + ... + 99 + 100) - X (1 + 2 + 3 + ... + 99 + 100) = 5050, SO WE CAN REPHRASE THIS TO SAY THAT THE NUMBER ON THE PAPER SHOULD EQUAL: Y = 5050 - X SO TO FIGURE OUT THE MISSING SERVANT'S NUMBER, THE KING SIMPLY NEEDS TO SUBTRACT THE NUMBER WRITTEN ON HIS PAPER FROM 5050: 5050 - Y = X IF THERE WAS ONLY ONE BLUE-EYED PERSON ON THE ISLAND, THEN THAT PERSON WOULD LOOK AROUND AND SEE THAT THERE IS NO OTHER BLUE-EYED PERSON. SO HE REALIZES THAT HE IS THE ONLY PERSON WITH BLUE EYES ON THE ISLAND AND LEAVES ON THE DAY OF THE ANNOUNCEMENT. IF THERE ARE 2 BLUE-EYED PEOPLE, THEN THEY LOOK AT EACH OTHER. EACH ONE EXPECTS THE OTHER TO LEAVE ON THE DAY OF THE ANNOUNCEMENT. HOWEVER, ON THE NEXT DAY, WHEN THEY REALIZE THAT NEITHER OF THEM LEFT THE ISLAND, THEY WOULD BE ABLE TO DEDUCE THAT BOTH OF THEM HAVE BLUE EYES. THEY BOTH LEAVE THE ISLAND ON THE SECOND DAY. THROUGH MATHEMATICAL INDUCTION, THIS LOGIC CAN BE APPLIED TO THE 100 BLUE-EYED PEOPLE ON THE ISLAND. SO ON THE 100TH DAY, ALL THE 100 BLUE-EYED PEOPLE LEAVE THE ISLAND THE GREEN ONE IS TELLING THE TRUTH. LETS ASSUME THAT ONE OF THEM IS TELLING THE TRUTH AND THEN TRY TO PROVE THAT. SINCE ALL FOUR SERVANTS ARE DISAGREEING THEN 3 OF THEM MUST BE LYING. THE SERVANT TELLING THE TRUTH WILL HAVE EITHER 6 OR 8 LEGS. THE OTHER 3 SERVANTS WILL HAVE 7 LEGS SINCE THEY LIE. SO THE TOTAL NUMBER OF LEGS SHOULD BE EITHER 27 (6 + 7 + 7 + 7) LEGS OR 29 (8 + 7 + 7 +7) LEGS. ONLY GREEN SERVANT COULD BE TELLING THE TRUTH AS IT SAID 27 LEGS. ALTERNATIVELY, LET'S SAY BLUE IS TELLING THE TRUTH: SO THE BLUE ONE HAS EITHER 6 OR 8 LEGS. AND EACH OF THE OTHER OCTOPUSES ARE LYING HENCE HAVE 7 LEGS EACH. SO OUR TOTAL NUMBER OF LEGS: 6 + 7 + 7 + 7 = 27 LEGS OR 8 + 7 + 7 + 7 = 29 LEGS. BUT SINCE BLUE SAID THAT ALTOGETHER THEY HAVE 28 LEGS, WE KNOW HE IS LYING. IF YOU FOLLOW THIS SAME LOGIC FOR ALL OF THEM, YOU REALIZE THAT ONLY THE GREEN OCTOPUS CAN BE TELLING THE TRUTH AS THE NUMBER OF LEGS ADDS UP