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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 THE ELDEST IS 9 YEARS OLD AND THE 2 YOUNGER ONES ARE 2 YEARS OLD. LET'S BREAK IT DOWN. THE PRODUCT OF THEIR AGES IS 36. SO THE POSSIBLE CHOICES ARE: 1,1,36 - SUM(1,1,36) = 38 1,6,6 - SUM(1,6,6) = 13 1,2,18 - SUM(1,2,18) = 21 1,3,12 - SUM(1,3,12) = 16 1,4,9 - SUM(1,4,9) = 14 2,2,9 - SUM(2,2,9) = 13 2,3,6 - SUM(2,3,6) = 11 3,3,4 - SUM(3,3,4) = 10 SIX OF THE SUMS ARE UNIQUE, SO IF IT WERE ONE OF THOSE, TOM WOULD HAVE RECOGNISED THE NUMBER ACROSS THE STREET THAT MATCHES AND HE WOULD KNOW THE ANSWER, BUT HE COULD NOT FIGURE OUT THE ANSWER. THIS MEANS THERE ARE TWO OR MORE COMBINATIONS WITH THE SAME SUM. FROM THE CHOICES ABOVE, ONLY TWO OF THEM ARE POSSIBLE NOW. 1,6,6 - SUM(1,6,6) = 13 2,2,9 - SUM(2,2,9) = 13 WHEN TOM HEARD THAT THE ELDEST IS VISITING HIS GRANDFATHER, WE CAN ELIMINATE COMBINATION 1 SINCE THERE ARE TWO ELDEST ONES. THIS LEAVES US WITH ONLY 1 OPTION LEFT, THAT IS 2, 2 AND 9 TOM WINS AGAIN. IN THE FIRST RACE TOM RAN 100 METERS IN THE TIME IT TOOK HARRY TO RUN 95 METERS. SO IN THE SECOND RACE WHEN HARRY IS AT THE 95 METER MARK TOM WILL ALSO BE THERE (SINCE 100 - 5 = 95). SINCE TOM IS FASTER HE WILL PASS HARRY IN THE LAST 5 METERS OF THE RACE THERE MAY MANY POSSIBLE SOLUTIONS, HERE IS A SIMPLE ONE: THE FIRST PERSON TAKES HIS MONTHLY SALARY (LETS SAY $100), ADDS AN ARBITRARY AMOUNT THAT ONLY HE KNOWS (LETS SAY $475) AND WHISPERS THAT NUMBER ($575) TO THE SECOND PERSON. THE SECOND PERSON THEN ADDS HIS MONTHLY SALARY (LETS SAY $150) TO THAT VALUE AND WHISPERS THE NEW TOTAL ($725) TO THE THIRD PERSON. THE THIRD PERSON ADDS IN HIS SALARY, WHISPERS THE NEW TOTAL TO THE FOURTH PERSON AND SO ON UNTIL THE TENTH PERSON ADDS IN HIS SALARY AND WHISPERS THE FINAL TOTAL TO THE FIRST PERSON. THE FIRST PERSON THEN SUBTRACTS HIS ARBITRARY NUMBER FROM THAT TOTAL, DIVIDES BY 10, WHICH PRODUCES THE AVERAGE MONTHLY SALARY FOR THE GROUP WITHOUT ANYONE HAVING TO DIVULGE THEIR SALARY TO ANYONE ELSE