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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 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 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 7 RACES. LET'S NAME THE RACES R1 THROUGH R7. LET RXN REPRESENT A HORSE IN RACE X, FINISHING IN NTH PLACE. SO R32 REPRESENTS A HORSE THAT FINISHED 2ND PLACE IN THE 3RD RACE. GROUP THE 25 HORSES INTO 5 GROUPS OF 5 AND RACE THEM. THE 4TH AND 5TH PLACED HORSES OF EACH RACE CAN BE ELIMINATED SINCE THEY CANNOT MEET THE CRITERIA OF 3 FASTEST HORSES. WE ARE NOW LEFT WITH 15 HORSES (5 GROUPS OF 3 HORSES); 3 HORSES FROM EACH RACE. FOR THE 6TH RACE, RACE THE FASTEST HORSE (R11, R21, R31, R41, R51) FROM EACH OF THE FIRST 5 RACES. THE WINNER OF THE 6TH RACE IS THE FASTEST HORSE. THE 4TH AND 5TH PLACED HORSES FROM THE 6TH RACE CAN BE ELIMINATED INCLUDING ALL THE HORSES WITHIN THEIR RESPECTIVE GROUPS. FOR EXAMPLE, IF THE HORSE THAT CAME IN 4TH PLACE IS FROM R41, THE HORSES R42 AND R43 CAN BE ELIMINATED AS WELL. LET'S SAY FOR THE 6TH RACE, R11 CAME FIRST, R21 CAME 2ND AND R31 CAME 3RD. WE KNOW THAT R11 IS THE FASTEST HORSE. WE NOW NEED TO DETERMINE THE 2ND AND 3RD FASTEST HORSES. WE CAN NOW ALSO ELIMINATE THE HORSES R23, R32 AND R33. THIS WILL LEAVE US WITH FIVE HORSES FOR THE 7TH RACE - R12, R13, R21, R22 AND R31. THE 1ST AND 2ND PLACED HORSES IN THE 7TH RACE ARE THE 2ND AND 3RD FASTEST HORSES