Y _ _
W _ L L
N _ _ D
T _
_ P _ N
_
M _ N _ M _ M
_ F
2
B _ X _ S .
F _ R S T ,
_ P _ N
T H _
B _ X
L _ B _ L L _ D
A P P L _ ;
_ F
_ T ' S
L _ B _ L L _ D
C _ R R _ C T L Y
W _ ' R _
D _ N _
( T H _ _ G H
T H _ S
_ S
P _ R _ L Y
B _ S _ D
_ N
L _ C K ) ,
_ T H _ R W _ S _
W _ ' L L
F _ N D
_ _ T H _ R
B _ N _ N _ S ,
C _ R R _ T S ,
_ R
D _ T _ S .
I N
_ N Y
_ F
T H _ S _
C _ S _ S ,
W _
K N _ W
_ _ T H _ R
T H _ T
T H _
B _ X _ S
L _ B _ L L _ D
B _ N _ N _ S ,
C _ R R _ T S ,
_ R
D _ T _ S
M _ S T
_ L S _
B _
M _ S L _ B _ L _ D .
I F
T H _
F _ R S T
_ P _ N _ D
B _ X
H _ S :
B _ N _ N _ S
_ N S _ D _ ,
T H _ N
T H _
B _ X
L _ B _ L L _ D
B _ N _ N _
_ S
_ N C _ R R _ C T
_ N D
_ _ T H _ R
C _ R R _ T S
_ R
D _ T _ S
_ R _
C _ R R _ C T .
C _ R R _ T S
_ N S _ D _ ,
T H _ N
T H _
B _ X
L _ B _ L L _ D
C _ R R _ T S
_ S
_ N C _ R R _ C T
_ N D
_ _ T H _ R
B _ N _ N _
_ R
D _ T _ S
_ R _
C _ R R _ C T
D _ T _ S
_ N S _ D _ ,
T H _ N
T H _
B _ X
L _ B _ L L _ D
D _ T _ S
_ S
_ N C _ R R _ C T
_ N D
_ _ T H _ R
B _ N _ N _
_ R
C _ R R _ T S
_ R _
C _ R R _ C T .
N _
M _ T T _ R
T H _
C _ R C _ M S T _ N C _ ,
_ F T _ R
_ P _ N _ N G
_ N _
B _ X
W _
C _ N
_ D _ N T _ F Y
T W _
R _ M _ _ N _ N G
B _ X _ S
T H _ T
M _ G H T
B _
T H _
C _ R R _ C T L Y
L _ B _ L L _ D
B _ X .
O P _ N
_ _ T H _ R
_ N _
_ F
T H _ M .
I F
_ T ' S
T H _
C _ R R _ C T L Y
L _ B _ L L _ D
B _ X ,
W _ ' V _
F _ _ N D
_ T .
O T H _ R W _ S _
T H _
R _ M _ _ N _ N G
_ N _ P _ N _ D
B _ X
_ S
C _ R R _ C T L Y
L _ B _ L L _ D
_ N D
W _ ' V _
F _ _ N D
_ T .
W _
D _ N ' T
N _ _ D
T _
_ P _ N
_ T
T _
D _ _ B L _
C _ N F _ R M Clue
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 YOU WOULD ONLY NEED TO TAKE OUT ONE MARBLE BECAUSE WE KNOW THAT ALL OF THE LABELS ARE INCORRECT. SO YOU PULL ONE MARBLE OUT OF THE BOX LABELED "MIXED." IF RED COMES OUT, YOU KNOW THAT HAS TO BE THE ALL-RED BOX, SO YOU PUT THE RED LABEL ON IT. THE BOX LABELLED "BLUE" MUST THEN BE LABELLED "MIXED" BECAUSE YOU KNOW IT IS ALSO LABELED INCORRECTLY, AND THEREFORE CAN'T BE BLUE. YOU WOULD LABEL THE LAST BOX "BLUE" BECAUSE THAT IS THE ONLY COLOR/BOX COMBO LEFT. IF THE FIRST MARBLE YOU PULLED OUT FROM THE BOX LABELED "MIXED" IS A BLUE MARBLE, THEN YOU SOLVE THE PROBLEM IN THE SAME GENERAL WAY THE ONE LABELED BOTH. SINCE YOU KNOW THAT THE BOX IS LABELED INCORRECTLY, IT MUST HAVE ALL BLACK MARBLES OR ALL WHITE MARBLES. IF A BLACK MARBLE COMES OUT, YOU KNOW THAT HAS TO BE THE BLACK MARBLES BOX, SO YOU PUT THE "BLACK" LABEL ON IT. THE BOX LABELLED "WHITE" MUST THEN BE LABELLED "BOTH" BECAUSE YOU KNOW IT IS ALSO LABELED INCORRECTLY, AND THEREFORE CAN'T BE WHITE. YOU WOULD LABEL THE LAST BOX "WHITE" BECAUSE THAT IS THE ONLY COLOR/BOX COMBO LEFT. IF THE FIRST MARBLE YOU PULLED OUT FROM THE BOX LABELED "BOTH" IS A WHITE MARBLE, THEN YOU SOLVE THE PROBLEM IN THE SAME GENERAL WAY YOU WILL NEED TO OPEN A MINIMUM OF 2 BOXES. FIRST, OPEN THE BOX LABELLED APPLE; IF IT'S LABELLED CORRECTLY WE'RE DONE (THOUGH THIS IS PURELY BASED ON LUCK), OTHERWISE WE'LL FIND EITHER BANANAS, CARROTS, OR DATES. IN ANY OF THESE CASES, WE KNOW EITHER THAT THE BOXES LABELLED BANANAS, CARROTS, OR DATES MUST ALSO BE MISLABELED. IF THE FIRST OPENED BOX HAS: BANANAS INSIDE, THEN THE BOX LABELLED BANANA IS INCORRECT AND EITHER CARROTS OR DATES ARE CORRECT. CARROTS INSIDE, THEN THE BOX LABELLED CARROTS IS INCORRECT AND EITHER BANANA OR DATES ARE CORRECT DATES INSIDE, THEN THE BOX LABELLED DATES IS INCORRECT AND EITHER BANANA OR CARROTS ARE CORRECT. NO MATTER THE CIRCUMSTANCE, AFTER OPENING ONE BOX WE CAN IDENTIFY TWO REMAINING BOXES THAT MIGHT BE THE CORRECTLY LABELLED BOX. OPEN EITHER ONE OF THEM. IF IT'S THE CORRECTLY LABELLED BOX, WE'VE FOUND IT. OTHERWISE THE REMAINING UNOPENED BOX IS CORRECTLY LABELLED AND WE'VE FOUND IT. WE DON'T NEED TO OPEN IT TO DOUBLE CONFIRM