1 4
_ S
T H _
L _ _ S T
N _ M B _ R
_ F
T R _ P S
T _
F _ N D
_ _ T
T H _
S _ L _ T _ _ N .
T H _
_ _ S _ _ S T
W _ Y
T _
D _
T H _ S
W _ _ L D
B _
T _
S T _ R T
F R _ M
T H _
F _ R S T
F L _ _ R
_ N D
D R _ P
T H _
C _ C _ N _ T .
I F
_ T
D _ _ S N ' T
B R _ _ K ,
M _ V _
_ N
T _
T H _
N _ X T
F L _ _ R .
I F
_ T
D _ _ S
B R _ _ K ,
T H _ N
W _
K N _ W
T H _
M _ X _ M _ M
F L _ _ R
T H _
C _ C _ N _ T
W _ L L
S _ R V _ V _ .
I F
W _
C _ N T _ N _ _
T H _ S
P R _ C _ S S ,
W _
W _ L L
_ _ S _ L Y
F _ N D
_ _ T
T H _
M _ X _ M _ M
F L _ _ R S
T H _
C _ C _ N _ T
W _ L L
S _ R V _ V _
W _ T H
J _ S T
_ N _
C _ C _ N _ T .
S _
T H _
M _ X _ M _ M
N _ M B _ R
_ F
T R _ _ S
_ S
1 0 0
F _ R
1 0 0
F L _ _ R S .
T H _ R _
_ S
_
B _ T T _ R
W _ Y .
L _ T ' S
S T _ R T
_ T
T H _
S _ C _ N D
F L _ _ R .
I F
T H _
C _ C _ N _ T
B R _ _ K S ,
T H _ N
W _
C _ N
_ S _
T H _
S _ C _ N D
C _ C _ N _ T
T _
G _
B _ C K
T _
T H _
F _ R S T
F L _ _ R
_ N D
T R Y
_ G _ _ N .
I F
T H _
1 S T
C _ C _ N _ T
D _ _ S
N _ T
B R _ _ K ,
T H _ N
W _
C _ N
G _
_ H _ _ D
_ N D
T R Y
_ N
T H _
4 T H
F L _ _ R
( _ N
M _ L T _ P L _ S
_ F
2 ) .
I F
_ T
_ V _ R
B R _ _ K S ,
S _ Y
_ T
F L _ _ R
N ,
T H _ N
W _
K N _ W
_ T
S _ R V _ V _ D
F L _ _ R
N - 2 .
T H _ T
L _ _ V _ S
_ S
W _ T H
J _ S T
F L _ _ R
N - 1
T _
T R Y
W _ T H
T H _
S _ C _ N D
C _ C _ N _ T .
W _ T H
T H _ S
M _ T H _ D ,
T H _
M _ X _ M _ M
T R _ _ S
_ S
5 1 .
I T
_ C C _ R S
W H _ N
T H _
C _ C _ N _ T
S _ R V _ V _ S
9 8
F L _ _ R S .
I T
W _ L L
T _ K _
5 0
T R _ _ S
T _
R _ _ C H
F L _ _ R
1 0 0
_ N D
_ N _
M _ R _
C _ C _ N _ T
T _
T R Y
_ N
T H _
9 9 T H
F L _ _ R
S _
T H _
T _ T _ L
_ S
5 1
T R _ _ S .
N _ W ,
F _ R
T H _
_ L T _ M _ T _
M _ T H _ D .
I N S T _ _ D
_ F
T _ K _ N G
_ Q _ _ L
_ N T _ R V _ L S ,
W _
C _ N
D _ C R _ _ S _
T H _
N _ M B _ R
_ F
F L _ _ R S
B Y
_ N _
L _ S S
T H _ N
T H _
P R _ V _ _ _ S
_ N _ .
F _ R
_ X _ M P L _ ,
L _ T ' S
F _ R S T
T R Y
_ T
F L _ _ R
1 4 .
I F
_ T
B R _ _ K S ,
T H _ N
W _
N _ _ D
1 3
M _ R _
T R _ _ S
T _
F _ N D
T H _
S _ L _ T _ _ N .
I F
_ T
D _ _ S N ' T
B R _ _ K ,
T H _ N
W _
S H _ _ L D
T R Y
F L _ _ R
2 7
( 1 4
+
1 3 ) .
I F
_ T
B R _ _ K S ,
W _
N _ _ D
1 2
M _ R _
T R _ _ S
T _
F _ N D
T H _
S _ L _ T _ _ N .
S _
T H _
_ N _ T _ _ L
2
T R _ _ S
P L _ S
T H _
_ D D _ T _ _ N _ L
1 2
T R _ _ S
W _ _ L D
S T _ L L
B _
1 4
T R _ _ S
_ N
T _ T _ L .
I F
_ T
D _ _ S N ' T
B R _ _ K ,
W _
C _ N
T R Y
3 9
( 2 7
+
1 2 )
_ N D
S _
_ N .
U S _ N G
1 4
_ S
T H _
_ N _ T _ _ L
F L _ _ R ,
W _
C _ N
R _ _ C H
_ P
T _
F L _ _ R
1 0 5
( 1 4
+
1 3
+
1 2
+
. . .
+
1 )
B _ F _ R _
W _
N _ _ D
M _ R _
T H _ N
1 4
T R _ _ S .
S _ N C _
W _
_ N L Y
N _ _ D
T _
C _ V _ R
1 0 0
F L _ _ R S ,
1 4
T R _ _ S
_ S
S _ F F _ C _ _ N T
T _
F _ N D
T H _
S _ L _ T _ _ N .
C _ C _ N _ T
D R _ P
C _ _ N T F L _ _ R
1 1 4
2 2 7
3 3 9
4 5 0
5 6 0
6 6 9
7 7 7
8 8 4
9 9 0
1 0 9 5
1 1 9 9
1 2 1 0 0
T H _ R _ F _ R _ ,
1 4
_ S
T H _
L _ _ S T
N _ M B _ R
_ F
T R _ _ S
T _
F _ N D
_ _ T
T H _
S _ L _ T _ _ N 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 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 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 14 IS THE LEAST NUMBER OF TRIPS TO FIND OUT THE SOLUTION. THE EASIEST WAY TO DO THIS WOULD BE TO START FROM THE FIRST FLOOR AND DROP THE COCONUT. IF IT DOESN'T BREAK, MOVE ON TO THE NEXT FLOOR. IF IT DOES BREAK, THEN WE KNOW THE MAXIMUM FLOOR THE COCONUT WILL SURVIVE. IF WE CONTINUE THIS PROCESS, WE WILL EASILY FIND OUT THE MAXIMUM FLOORS THE COCONUT WILL SURVIVE WITH JUST ONE COCONUT. 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 COCONUT BREAKS, THEN WE CAN USE THE SECOND COCONUT TO GO BACK TO THE FIRST FLOOR AND TRY AGAIN. IF THE 1ST COCONUT 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 COCONUT. WITH THIS METHOD, THE MAXIMUM TRIES IS 51. IT OCCURS WHEN THE COCONUT SURVIVES 98 FLOORS. IT WILL TAKE 50 TRIES TO REACH FLOOR 100 AND ONE MORE COCONUT 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. COCONUT 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