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THE MAXIMUM NUMBER OF BANANAS THAT CAN BE TRANSFERRED IS 533. IF WE TRANSPORT 1000 BANANAS AT A TIME, THE CAMEL WILL CONSUME ALL THE BANANAS BY THE TIME IT REACHES THE DESTINATION. SO, WE NEED TO HAVE INTERMEDIATE DROP POINTS, THE CAMEL CAN THEN MAKE SEVERAL SHORT TRIPS IN BETWEEN. TO BE OPTIMAL, WE TRY TO MAINTAIN THE NUMBER OF BANANAS AT EACH POINT TO BE A MULTIPLE OF 1000, AS THAT'S THE MAXIMUM OF BANANAS THE CAMEL CAN TRANSPORT AT ANY POINT OF TIME. SOURCE---IP1---IP2----DESTINATION 3000 X KM 2000 Y KM 1000 Z KM TO GO FROM SOURCE TO IP1 POINT CAMEL HAS TO TAKE A TOTAL OF 5 TRIPS, 3 FORWARD AND 2 BACKWARDS, SINCE WE HAVE 3000 BANANAS TO TRANSPORT. THE SAME WAY FROM IP1 TO IP2 CAMEL HAS TO TAKE A TOTAL OF 3 TRIPS, 2 FORWARD AND 1 BACKWARD, SINCE WE HAVE 2000 BANANAS TO TRANSPORT. FROM IP2 TO DESTINATION WE ONLY HAVE 1 FORWARD MOVE. LET'S SEE THE TOTAL NUMBER OF BANANAS CONSUMED AT EVERY POINT. FROM THE SOURCE TO IP1 ITS 5X BANANAS, AS THE DISTANCE BETWEEN THE SOURCE AND IP1 IS X KM AND THE CAMEL HAD 5 TRIPS. FROM IP1 TO IP2 ITS 3Y BANANAS, AS THE DISTANCE BETWEEN IP1 AND IP2 IS Y KM AND THE CAMEL HAD 3 TRIPS. FROM IP2 TO DESTINATION ITS Z BANANAS. WE CAN NOW CALCULATE THE DISTANCE BETWEEN THE POINTS: 3000 - 5X = 2000 SO WE GET X = 200 2000-3Y = 1000 SO WE GET Y = 333.33 BUT HERE THE DISTANCE IS ALSO THE NUMBER OF BANANAS AND IT CANNOT BE FRACTION SO WE TAKE Y = 333 AND AT IP2 WE HAVE THE NUMBER OF BANANAS EQUAL 1001, SO ITS 2000-3Y = 1001 SO THE REMAINING DISTANCE TO THE MARKET IS 1000 - X - Y = Z I.E 1000-200-333 = Z = 467. FROM IP2 TO THE DESTINATION POINT, THE CAMEL CONSUMES 467 BANANAS AND 533 BANANAS REMAIN. REFERENCE: A CAMEL TRANSPORTING BANANAS - PUZZLING STACK EXCHANGE 50 STATEMENTS ARE CORRECT AND 50 ARE INCORRECT. STATEMENTS 1 TO 50 ARE TRUE AND STATEMENTS 51 TO 100 ARE FALSE. IF ANY ONE OF THE STATEMENTS IS TRUE, THEN ALL OF THE STATEMENTS NUMBERED LOWER THAN THAT ONE MUST ALSO BE TRUE, BECAUSE THE TERM "AT LEAST" IS INCLUSIVE OF A FEWER NUMBER, AND THEY ARE NUMBERED IN ASCENDING ORDER. ASSUMING STATEMENT 1 IS TRUE; THEN STATEMENT 100 IS FALSE. ASSUMING STATEMENTS 1-2 ARE TRUE; THEN STATEMENTS 99-100 ARE FALSE. ASSUMING STATEMENTS 1-3 ARE TRUE; THEN STATEMENTS 98-100 ARE FALSE. ASSUMING STATEMENTS 1-4 ARE TRUE; THEN STATEMENTS 97-100 ARE FALSE. ... ASSUMING STATEMENTS 1-50 ARE TRUE; THEN STATEMENTS 51-100 ARE FALSE. STATEMENT 99 WILL BE THE ONLY CORRECT STATEMENT IF THE TERMS "AT LEAST" ARE REPLACED WITH THE TERM "EXACTLY.", E.G. "EXACTLY 99 OF THESE STATEMENTS ARE FALSE." THE DAUGHTER SHOULD PICK ENVELOPE 1. STATEMENTS 1 AND 2 ARE FALSE, AND THE ONLY TRUE STATEMENT IS STATEMENT 3. IF THE CHECK WAS IN ENVELOPE 1, THAT WOULD MAKE STATEMENT 1 AND STATEMENT 2 FALSE AND STATEMENT 3 WOULD BE THE ONLY TRUE STATEMENT. IF THE CHECK WAS IN ENVELOPE 2, BOTH STATEMENTS 1 AND 2 WOULD BE TRUE. IF THE CHECK WAS IN ENVELOPE 3, BOTH STATEMENTS 1 AND 3 WOULDBE TRUE THE MINIMUM NUMBER OF WEIGHTS REQUIRED IS FIVE AND THESE SHOULD WEIGHT 1, 3, 9, 27 AND 81 POUNDS. THE MERCHANT HAS TO USE A BALANCE WEIGHING SCALE TO DO THE JOB. TO WEIGH 2 POUNDS, HE'LL HAVE TO PUT THE 3 POUND WEIGHT ON ONE PAN AND 1 POUND WEIGHT ON THE OTHER PAN. TO WEIGH 5 POUNDS, HE'LL HAVE TO PUT THE 9 POUND WEIGHT ON ONE PAN AND 1 AND 3 POUND WEIGHTS ON THE OTHER PAN