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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 FIRST CUT THE CAKE INTO 4 EQUAL PIECES WITH 2 CUTS ON THE TOP - ONE HORIZONTALLY DOWN THE CENTER OF THE CAKE AND THE OTHER VERTICALLY DOWN THE CENTER OF THE CAKE LIKE A CROSS ('+'). NEXT IS TO CUT THE 4 PIECES INTO HALF WITH A HORIZONTAL FINAL CUT ON THE SIDE. THIS WILL GIVE YOU 4 PIECES OF CAKE ON TOP AND 4 PIECES OF CAKE UNDERNEATH TAKE ONE COIN FROM THE FIRST BAG, TWO FROM THE SECOND, THREE FROM THE THIRD, AND SO ON, UNTIL YOU HAVE TEN COINS FROM THE TENTH. THEN STACK AND WEIGH THE 55 COINS (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10). SINCE EACH GOLD COIN WEIGH TEN GRAMS, IF ALL HAD BEEN GOLD, THE SCALE WOULD HAVE READ 550 GRAMS. THE AMOUNT BY WHICH THE WEIGHT WAS TOO LIGHT INDICATED THE NUMBER OF SILVER COINS AND THE NUMBER OF THE THE SILVER BAG. FOR INSTANCE, IF THE WEIGHT WAS 543 GRAMS, IT WOULD INDICATE THAT 7 SILVER COINS (550 - 543 = 7) HAD BEEN WEIGHED WITH THE GOLD COINS AND THAT THE REST OF THE SILVER COINS WERE IN THE SEVENTH BAG SHE WILL HAVE FIVE PIECES OF CHAINS AFTER THE 2 CUTS. THE CHAINS WILL HAVE LOOPS OF 1, 1, 3, 6 AND 12. THE CHAIN IS MADE UP OF INTERLOCKING LOOPS. SO WHEN YOU MAKE 1 CUT IN THE CHAIN IN THE MIDDLE, YOU WOULD END UP WITH 2 CHAINS AND 1 CUT LOOP. SO 2 CUTS IN THE MIDDLE WOULD LEAVE YOU WITH 3 CHAINS AND 2 CUT SINGLE LOOPS. SO THE JEWELER MAKES ONE CUT ON THE 4TH LOOP MAKING A 1 CUT LOOP, A 3 LOOP CHAIN, AND A 19 LOOP CHAIN. THEN A CUT ON THE 7TH LOOP OF THE 19 LOOP CHAIN LEAVING A 1 CUT LOOP, A 6 LOOP CHAIN, AND A 12 LOOP CHAIN. SO AT THE END SHE HAS 1 CUT LOOP, ANOTHER 1 CUT LOOP, A 3 LOOP CHAIN, A 6 LOOP CHAIN, AND A 12 LOOP CHAIN