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_ N _ Q _ _
C _ M B _ N _ T _ _ N S Clue
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 THE BIN WAS HALF FULL AT 10PM. THE SKIP STARTED WITH 1 LONELY BOTTLE. AT NOON: 1 PERSON CAME ALONG AND ADDED A BOTTLE, MAKING THE TOTAL 2 BOTTLES. AT 1PM: 2 PEOPLE CAME ALONG AND ADDED A BOTTLE EACH, MAKING THE TOTAL 2 + 2 = 4 BOTTLES. AT 2PM: 4 PEOPLE CAME ALONG AND ADDED A BOTTLE EACH, MAKING THE TOTAL 4 + 4 = 8 BOTTLES. AT 3PM: 8 PEOPLE CAME ALONG AND ADDED A BOTTLE EACH, MAKING THE TOTAL 8 + 8 = 16 BOTTLES. THEREFORE THE NUMBER OF BOTTLES IN THE BIN IS DOUBLING EVERY HOUR, AS IT WAS FULL AT 11PM, IT MUST HAVE BEEN HALF FULL AT 10PM THE KING KEEPS POISON 12 SO THAT HE CAN NEUTRALIZE ANY POISON THAT THE CLOWN GIVES HIM. HE POURS POISON 10 (HIS SECOND STRONGEST POISON) AND GIVES IT TO THE CLOWN TO DRINK. CLOWN CAN DEDUCE THIS SO HE KEEPS POISON 11 FOR HIMSELF TO NEUTRALIZE POISON 10. THAT'S HOW HE SURVIVES. HE ALSO KNOWS THAT THE KING, AFTER DRINKING THE CUP HE GAVE HIM, WILL DRINK POISON 12 TO NEUTRALIZE THE POISON. SO THE CLOWN POURS PLAIN WATER INSTEAD OF POISON INTO THE CUP FOR THE KING. THE KING DRINKS THE WATER (WATER IS NOT POISON) AND RIGHT AFTER THAT HE DRINKS THE STRONGEST POISON POSSIBLE (12) AND DIES THE JAILOR NEEDS 10 PRISONERS. BINARY MATHS IS NEEDED TO SOLVE THIS PUZZLE. BELOW IS AN EXAMPLE ON HOW YOU USE BINARY LOGIC TO FIND THE NUMBER OF PRISONERS FOR 8 BOTTLES OF WINE. ASSUME THE WINE BOTTLES ARE NAMED W1, W2, W3...W8. THE PRISONERS ARE NAMED P1, P2 AND P3. THE ABOVE CHART SUMMARISES WHICH PRISONER HAS TO DRINK FROM WHICH WINE BOTTLE. '1' INDICATES THAT THE PRISONER HAS TO DRINK FROM THAT BOTTLE. BOTTLE W1 IS NOT FED TO ANY PRISONER. BOTTLE W2 IS FED TO PRISONER P3. BOTTLE W3 IS FED TO PRISONER P2 AND SO ON. IF NO ONE DIES, THEN WINE BOTTLE W1 IS POISONED. IF ONLY PRISONER P3 DIES, BOTTLE W2 IS POISONED. IF ONLY PRISONER P2 DIES, BOTTLE W3 IS POISONED. IF BOTH PRISONERS P2 AND P3 DIE, BOTTLE W4 IS POISONED. IF ONLY PRISONER P1 DIES, BOTTLE W5 IS POISONED. IF BOTH PRISONERS P1 AND P3 DIE, BOTTLE W6 IS POISONED. IF BOTH PRISONERS P1 AND P2 DIE, BOTTLE W7 IS POISONED. IF ALL 3 PRISONERS DIE, BOTTLE W8 IS POISONED. SO TO TEST 1000 BOTTLES OF WINE, 10 PRISONERS ARE SUFFICIENT AS THAT WILL ALLOW (2^10) 1024 UNIQUE COMBINATIONS