T H _
S T _ D _ N T
_ S
D _ _ B L _
C _ _ N T _ N G
_
L _ T
_ F
T H _
D _ Y S .
A
L _ T
_ F
T H _
T _ M _
S P _ N T
S L _ _ P _ N G ,
_ _ T _ N G ,
_ N D
R _ L _ X _ N G
_ C C _ R S
D _ R _ N G
W _ _ K _ N D S
_ N D
T H _
S _ M M _ R .
W _ _ K _ N D S
_ L S _
_ C C _ R
D _ R _ N G
T H _
S _ M M _ R ,
S _
_ L L
_ F
T H _ S _
H _ _ R S
_ R _
G _ T T _ N G
C _ _ N T _ D
S _ V _ R _ L
T _ M _ S .
A N D ,
S C H _ _ L
_ S
N _ T
_ N
_ L L
D _ Y
_ F F _ _ R .
S _
T H _
4
D _ Y S
_ C T _ _ L L Y
R _ P R _ S _ N T S
M _ R _
D _ Y S
_ F
S C H _ _ L .
I F
S C H _ _ L
_ S
6
H _ _ R S
P _ R
D _ Y ,
T H _ S _
F _ _ R
D _ Y S
R _ P R _ S _ N T S
1 6
D _ Y S
_ F
S C H _ _ L Clue
7 DAYS. ONE COP HAS TO SEARCH CLOCKWISE AND THE OTHER COP HAS TO SEARCH ANTI CLOCKWISE. SO, THE COPS START SEARCHING AT: CAVE 13 AND CAVE 1 ON THE 1ST DAY CAVE 12 AND CAVE 2 ON 2ND DAY CAVE 11 AND CAVE 3 ON 3RD DAY CAVE 10 AND CAVE 4 ON 4TH DAY CAVE 9 AND CAVE 5 ON 5TH DAY CAVE 8 AND CAVE 6 ON 6TH DAY CAVE 7 ON 7TH DAY THE WORST CASE IS WHEN THE THIEF STAYS IN CAVE 7 AND DOES NOT MOVE THEIR HATS ARE ALL BLUE IN COLOR. THERE ARE 3 POSSIBLE HAT COLOR COMBINATIONS: [A] 1 BLUE, 2 WHITE [B] 2 BLUE, 1 WHITE [C] 3 BLUE THE COLOR COMBINATION OF 3 WHITE HATS IS NOT POSSIBLE SINCE THE KING HAS ALREADY SAID THAT AT LEAST ONE OF THE WISE MEN HAS A BLUE HAT. SO, LET'S START OUR ANALYSIS. WHAT IF THERE WERE ONE BLUE HAT AND TWO WHITE HATS? THEN THE WISE MAN WITH THE BLUE HAT WOULD HAVE SEEN TWO WHITE HATS AND IMMEDIATELY CALLED OUT THAT HIS OWN HAT WAS BLUE, SINCE HE KNEW THERE IS AT LEAST ONE BLUE HAT. THIS DIDN'T HAPPEN, AND THUS THE HAT COLOR COMBINATION [A] IS RULED OUT. NOW, THE WISE MEN KNEW THAT ONLY 2 HAT COLOR COMBINATIONS ARE POSSIBLE - COMBINATION [B] OR [C]. WHAT IF THERE WERE TWO BLUE HATS AND ONE WHITE HAT? THE MEN WITH THE BLUE HATS WILL SEE ONE WHITE HAT AND ONE BLUE HAT. THEY WILL CONCLUDE THAT COLOR COMBINATION [B] IS THE CASE AND WOULD CALL OUT BLUE AS THEIR HAT COLOR. THIS ALSO DID NOT HAPPEN, AND THUS THE HAT COLOR COMBINATION [B] IS ALSO RULED OUT. AFTER SOME TIME, WHEN NONE OF THE WISE MEN ARE ABLE TO IDENTIFY THE COLOR OF THEIR OWN HATS, COMBINATION [C] (OF 3 BLUE HATS) BECOMES THE ONLY POSSIBLE OPTION WHEN THE FIRST SERVANT COMES IN, THE KING SHOULD WRITE DOWN HIS NUMBER. FOR EACH OTHER SERVANT THAT REPORTS IN, THE KING SHOULD ADD THAT SERVANT'S NUMBER TO THE CURRENT NUMBER WRITTEN ON THE PAPER, AND THEN WRITE THIS NEW NUMBER ON THE PAPER. LET X BE THE NUMBER OF THE MISSING SERVANT AND Y BE THE NUMBER THAT THE KING HAS WRITTEN. ONCE THE FINAL SERVANT HAS REPORTED IN, THE NUMBER ON THE PAPER SHOULD EQUAL: Y = (1 + 2 + 3 + ... + 99 + 100) - X (1 + 2 + 3 + ... + 99 + 100) = 5050, SO WE CAN REPHRASE THIS TO SAY THAT THE NUMBER ON THE PAPER SHOULD EQUAL: Y = 5050 - X SO TO FIGURE OUT THE MISSING SERVANT'S NUMBER, THE KING SIMPLY NEEDS TO SUBTRACT THE NUMBER WRITTEN ON HIS PAPER FROM 5050: 5050 - Y = X THE STUDENT IS DOUBLE COUNTING A LOT OF THE DAYS. A LOT OF THE TIME SPENT SLEEPING, EATING, AND RELAXING OCCURS DURING WEEKENDS AND THE SUMMER. WEEKENDS ALSO OCCUR DURING THE SUMMER, SO ALL OF THESE HOURS ARE GETTING COUNTED SEVERAL TIMES. AND, SCHOOL IS NOT AN ALL DAY AFFAIR. SO THE 4 DAYS ACTUALLY REPRESENTS MORE DAYS OF SCHOOL. IF SCHOOL IS 6 HOURS PER DAY, THOSE FOUR DAYS REPRESENTS 16 DAYS OF SCHOOL