H _
H _ D
1 5
_ G G S .
T H _
F _ R S T
P _ R S _ N
B _ _ G H T
H _ L F
_ F
T H _
1 5
_ G G S
W H _ C H
_ S
7 ½
_ G G S
_ N D
_ N _ T H _ R
½
_ G G .
S _
H _
B _ _ G H T
8
_ G G S
_ N
T _ T _ L
_ N D
J _ C K
_ S
L _ F T
W _ T H
7
_ G G S .
T H _
S _ C _ N D
P _ R S _ N
B _ _ G H T
H _ L F
_ F
T H _
R _ M _ _ N _ N G
7
_ G G S
W H _ C H
_ S
3 ½
_ G G S
_ N D
_ N _ T H _ R
½
_ G G .
S _
H _
B _ _ G H T
4
_ G G S
_ N D
J _ C K
_ S
L _ F T
W _ T H
3
_ G G S .
T H _
L _ S T
P _ R S _ N
B _ _ G H T
H _ L F
_ F
T H _
R _ M _ _ N _ N G
3
_ G G S
W H _ C H
_ S
1 ½
_ G G S
_ N D
_ N _ T H _ R
½
_ G G .
S _
T H _
L _ S T
P _ R S _ N
B _ _ G H T
2
_ G G S .
T H _ T
L _ _ V _ S
1
_ G G
R _ M _ _ N _ N G
_ N
J _ C K ' S
B _ S K _ T Clue
HE HAD 15 EGGS. THE FIRST PERSON BOUGHT HALF OF THE 15 EGGS WHICH IS 7½ EGGS AND ANOTHER ½ EGG. SO HE BOUGHT 8 EGGS IN TOTAL AND JACK IS LEFT WITH 7 EGGS. THE SECOND PERSON BOUGHT HALF OF THE REMAINING 7 EGGS WHICH IS 3½ EGGS AND ANOTHER ½ EGG. SO HE BOUGHT 4 EGGS AND JACK IS LEFT WITH 3 EGGS. THE LAST PERSON BOUGHT HALF OF THE REMAINING 3 EGGS WHICH IS 1½ EGGS AND ANOTHER ½ EGG. SO THE LAST PERSON BOUGHT 2 EGGS. THAT LEAVES 1 EGG REMAINING IN JACK'S BASKET 25 EGGS WERE BROKEN. THERE IS ONLY ONE WAY OF FINDING A SOLUTION TO THIS PROBLEM: NUMBERS WHICH LEAVE A REMAINDER OF 1, WHEN DIVIDED BY 2: 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35 NUMBERS WHICH LEAVE A REMAINDER OF 1, WHEN DIVIDED BY 3: 4, 7, 10, 13, 16, 19, 22, 25, 28, 31, 34, 37 NUMBERS WHICH LEAVE A REMAINDER OF 1, WHEN DIVIDED BY 4: 5, 9, 13, 17, 21, 25, 29, 33, 37 NUMBERS WHICH LEAVE NO REMAINDER WHEN DIVIDED BY 5: 5, 10, 15, 20, 25, 30, 35 THE ONLY NUMBER FULFILLING THE FOUR CONDITIONS IS 25 7 EGGS. THE FIRST PERSON BOUGHT ONE HALF OF HIS EGGS PLUS ONE HALF AN EGG (3 1/2 + 1/2 = 4 EGGS) THIS LEFT HIM 3 EGGS. THE SECOND PERSON BOUGHT ONE-HALF OF HIS EGGS PLUS ONE HALF AN EGG, (1 1/2 + 1/2 = 2 EGGS) LEAVING THE MAN 1 EGG. THE LAST PERSON BOUGHT ONE-HALF OF HIS EGGS PLUS ONE-HALF AN EGG, (1/2 + 1/2 = 1 EGG) LEAVING NO EGGS 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