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1 0 P M Clue
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 KOLKHOZ IS 120 MILES AWAY FROM THE CITY AND THE TRUCK SHOULD TRAVEL AT 24 MILES AN HOUR. AT 30 MILES PER HOUR A TRUCK TRAVELS A MILE IN 2 MINUTES; AT 20 MILES PER HOUR, IN 3 MINUTES. AT THE LATTER SPEED THE TRUCK IS 1 MINUTE SLOWER PER MILE. TO LOSE 2 HOURS, OR 120 MINUTES, TAKES 120 MILES, WHICH IS HOW FAR THE KOLHOZ IS FROM THE CITY. AT 30 MPH, THE TRUCK WOULD COVER 120 MILES IN 4 HOURS. THE TRIP IS TO TAKE 1 HOUR LONGER, OR TO ARRIVE AT 11:00 A.M., AND CALLS FOR A SPEED OF 24 (120/5) MILES PER HOUR USE 4 HALF FILLED BARRELS TO FILL 2 EMPTY BARRELS. WE ARE THEN LEFT WITH 9 FULLY FILLED BARRELS, 3 HALF FILLED BARRELS AND 9 EMPTY BARRELS. EACH SON GETS 3 FULL BARRELS, 1 HALF-FULL BARREL, AND 3 EMPTY BARRELS 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