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E X C H _ N G _ Clue
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 CHILDREN ARE 1, 6 AND 6 YEARS OLD. THE PRODUCT OF THEIR AGES IS 36, SO NONE OF THEM CAN BE OLDER THAN 36. THE NUMBER 36 HAS TO BE EXPRESSED AS THE PRODUCT OF 3 NUMBERS. THEIR POSSIBLE AGES ARE (THE SUM OF THEIR AGES IS IN BRACKETS): 1, 1, 36 (3938) 1, 2, 18 (21) 1, 3, 12 (16) 1, 4, 9 (14) 1, 6, 6 (13) 2, 2, 9 (13) 2, 3, 6 (11) 3, 3, 4 (10) SINCE CHERYL IS TOM'S NEXT DOOR NEIGHBOUR, TOM KNOWS CHERYL'S HOUSE NUMBER. TOM WOULD KNOW THE CHILDREN'S AGES IN EVERY CASE THAT SUMS UP TO A UNIQUE NUMBER EXCEPT FOR THE SUM OF 13, WHICH HAVE 2 COMBINATIONS OF POSSIBLE AGES. AS A RESULT, TOM WOULD BE CONFUSED AS HE HAS TO PICK BETWEEN THE 2 COMBINATIONS: (1,6,6) AND (2,2,9). CHERYL THEN TELLS TOM ABOUT HER YOUNGEST CHILD WHO LIKES STRAWBERRY MILK WHICH TELLS TOM THAT THERE IS ONLY 1 YOUNGEST CHILD 14 IS THE LEAST NUMBER OF TRIES TO FIND OUT THE SOLUTION. THE EASIEST WAY TO DO THIS WOULD BE TO START FROM THE FIRST FLOOR AND DROP THE EGG. IF IT DOESN'T BREAK, MOVE ON TO THE NEXT FLOOR. IF IT DOES BREAK, THEN WE KNOW THE MAXIMUM FLOOR THE EGG WILL SURVIVE. IF WE CONTINUE THIS PROCESS, WE WILL EASILY FIND OUT THE MAXIMUM FLOORS THE EGG WILL SURVIVE WITH JUST ONE EGG. SO THE MAXIMUM NUMBER OF TRIES IS 100 FOR 100 FLOORS. THERE IS A BETTER WAY. LET'S START AT THE SECOND FLOOR. IF THE EGG BREAKS, THEN WE CAN USE THE SECOND EGG TO GO BACK TO THE FIRST FLOOR AND TRY AGAIN. IF THE 1ST EGG DOES NOT BREAK, THEN WE CAN GO AHEAD AND TRY ON THE 4TH FLOOR (IN MULTIPLES OF 2). IF IT EVER BREAKS, SAY AT FLOOR N, THEN WE KNOW IT SURVIVED FLOOR N-2. THAT LEAVES US WITH JUST FLOOR N-1 TO TRY WITH THE SECOND EGG. WITH THIS METHOD, THE MAXIMUM TRIES IS 51. IT OCCURS WHEN THE EGG SURVIVES 98 FLOORS. IT WILL TAKE 50 TRIES TO REACH FLOOR 100 AND ONE MORE EGG TO TRY ON THE 99TH FLOOR SO THE TOTAL IS 51 TRIES. NOW, FOR THE ULTIMATE METHOD. INSTEAD OF TAKING EQUAL INTERVALS, WE CAN DECREASE THE NUMBER OF FLOORS BY ONE LESS THAN THE PREVIOUS ONE. FOR EXAMPLE, LET'S FIRST TRY AT FLOOR 14. IF IT BREAKS, THEN WE NEED 13 MORE TRIES TO FIND THE SOLUTION. IF IT DOESN'T BREAK, THEN WE SHOULD TRY FLOOR 27 (14 + 13). IF IT BREAKS, WE NEED 12 MORE TRIES TO FIND THE SOLUTION. SO THE INITIAL 2 TRIES PLUS THE ADDITIONAL 12 TRIES WOULD STILL BE 14 TRIES IN TOTAL. IF IT DOESN'T BREAK, WE CAN TRY 39 (27 + 12) AND SO ON. USING 14 AS THE INITIAL FLOOR, WE CAN REACH UP TO FLOOR 105 (14 + 13 + 12 + ... + 1) BEFORE WE NEED MORE THAN 14 TRIES. SINCE WE ONLY NEED TO COVER 100 FLOORS, 14 TRIES IS SUFFICIENT TO FIND THE SOLUTION. EGG DROP COUNTFLOOR 114 227 339 450 560 669 777 884 990 1095 1199 12100 THEREFORE, 14 IS THE LEAST NUMBER OF TRIES TO FIND OUT THE SOLUTION