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IN 60 DAYS. IF ONE CLOCK GAINS A MINUTE A DAY (OR LOSES, THE MATH WILL BE THE SAME), IT WILL GAIN 24 MINUTES THE FIRST DAY, 48 MINUTES BY THE END OF THE SECOND, AND 120 MINUTES AFTER 5 DAYS. THIS MEANS IN TEN DAYS IT WILL GAIN 4 HOURS AND IN 20 DAYS, 8 HOURS. THIS TIMES 3, TO MAKE IT 24 HOURS, WILL REQUIRE 60 DAYS. THE OTHER CLOCK RUNNING BACKWARD WILL TELL THE SAME TIME AS THE NORMAL CLOCK EVERY 24 HOURS, SO IT REALLY DOESN'T PRESENT A PROBLEM FOR THE SOLUTION OF THE PUZZLE LET: CAMPER 1 = C1 (1) CAMPER 2 = C2 (2) CAMPER 3 = C3 (5) CAMPER 4 = C4 (10) THIS IS HOW THEY CROSS THE BRIDGE: C1,C2 → (2) ← C2 (2) C3,C4 → (10) ← C1 (1) C1,C2 → (2) THE TIME IN BRACKETS () INDICATE THE TIME TAKEN TO CROSS THE BRIDGE. IF YOU SUM THEM UP, IT WILL ADD UP TO 17 MINUTES THE TRAIN IS ABOUT TO LEAVE AND YOU WILL NOT BE ABLE TO CATCH THE TRAIN. THE AMOUNT OF TIME TAKEN TO TRAVEL TO THE HALFWAY POINT AT 15 MILES AN HOUR IS THE SAME AMOUNT OF TIME TAKEN TO REACH THE TRAIN STATION WHEN TRAVELLING TO THE RAILROAD STATION AT 30 MILES AN HOUR. FOR EXAMPLE, LET'S ASSUME THAT THE DISTANCE TO TRAVEL IS 30 MILES. WHEN TRAVELLING AT 30 MILES AN HOUR, IT TAKES 1 HOUR. WHEN TRAVELLING AT 15 MILES AN HOUR, IT WILL ALREADY TAKE 1 HOUR TO REACH THE HALFWAY POINT AT 3HR 16 MIN 21.82 (SOMEWHERE BETWEEN 21 AND 22 SECONDS) SECONDS. BY THE TIME THE MINUTE HAND GETS TO 3, THE HOUR HAND WOULD HAVE MOVED A LITTLE, SO THEY WILL NOT BE EXACTLY ON TOP OF EACH OTHER AT 3.15PM. THE MINUTE HAND AND HOUR HAND MOVE 360 DEGREES AND 30 DEGREES RESPECTIVELY IN 60 MINUTES