T H _ R _
_ R _
6
' F ' S
_ N
T H _
S _ N T _ N C _ .
O N _
_ F
_ V _ R _ G _
_ N T _ L L _ G _ N C _
F _ N D S
3
_ F
T H _ M .
I F
Y _ _
S P _ T T _ D
4 ,
Y _ _ ' R _
_ B _ V _
_ V _ R _ G _ .
T _
S _ _
5
_ S
R _ R _ .
I F
Y _ _
F _ _ N D
6 ,
Y _ _
_ R _
P R _ B _ B L Y
_
G _ N _ _ S .
T H _ R _
_ S
N _
C _ T C H .
M _ S T
P _ _ P L _
F _ R G _ T
T H _
' O F ' S .
T H _
H _ M _ N
B R _ _ N
T _ N D S
T _
S _ _
T H _ M
_ S
' V '
_ N S T _ _ D
_ F
' F ' Clue
THERE ARE 40 HOUSES. THERE ARE 16 HOUSES BETWEEN NUMBER 12 AND NUMBER 29. SINCE HALF OF THOSE HAVE TO BE ON EACH SIDE, THERE ARE 8 MORE HOUSES ON EACH SIDE. THIS MAKES THE LAST HOME ON ONE SIDE HOUSE NUMBER 20, AND THERE MUST BE 20 MORE HOMES GOING BACK UP THE STREET, WHICH MAKES»A TOTAL OF 40 THERE ARE 6 'F'S IN THE SENTENCE. ONE OF AVERAGE INTELLIGENCE FINDS 3 OF THEM. IF YOU SPOTTED 4, YOU'RE ABOVE AVERAGE. TO SEE 5 IS RARE. IF YOU FOUND 6, YOU ARE PROBABLY A GENIUS. THERE IS NO CATCH. MOST PEOPLE FORGET THE 'OF'S. THE HUMAN BRAIN TENDS TO SEE THEM AS 'V' INSTEAD OF 'F' PAT MUST HAVE PAINTED SIX MORE LAMP POSTS THAN TIM, NO MATTER HOW MANY LAMP POSTS THERE WERE. SUPPOSE THERE WERE 12 LAMP POSTS ON EACH SIDE; THEN PAT PAINTED 15 AND TIM 9. IF THERE WERE 100 LAMP POSTS ON EACH SIDE, PAT PAINTED 103, AND TIM ONLY 97 IF THERE WAS ONLY ONE BLUE-EYED PERSON ON THE ISLAND, THEN THAT PERSON WOULD LOOK AROUND AND SEE THAT THERE IS NO OTHER BLUE-EYED PERSON. SO HE REALIZES THAT HE IS THE ONLY PERSON WITH BLUE EYES ON THE ISLAND AND LEAVES ON THE DAY OF THE ANNOUNCEMENT. IF THERE ARE 2 BLUE-EYED PEOPLE, THEN THEY LOOK AT EACH OTHER. EACH ONE EXPECTS THE OTHER TO LEAVE ON THE DAY OF THE ANNOUNCEMENT. HOWEVER, ON THE NEXT DAY, WHEN THEY REALIZE THAT NEITHER OF THEM LEFT THE ISLAND, THEY WOULD BE ABLE TO DEDUCE THAT BOTH OF THEM HAVE BLUE EYES. THEY BOTH LEAVE THE ISLAND ON THE SECOND DAY. THROUGH MATHEMATICAL INDUCTION, THIS LOGIC CAN BE APPLIED TO THE 100 BLUE-EYED PEOPLE ON THE ISLAND. SO ON THE 100TH DAY, ALL THE 100 BLUE-EYED PEOPLE LEAVE THE ISLAND