4 0 0
M _ L _ S .
W _ R K _ N G
B _ C K W _ R D S ,
_ N
T H _
F _ N _ L
D _ Y
T _ M
C Y C L _ D
T H _
R _ M _ _ N _ N G
5
M _ L _ S .
O N
D _ Y
5 ,
H _
S T _ R T _ D
W _ T H
1 5
M _ L _ S
( C Y C L _ D
T W _
T H _ R D S ,
L _ _ V _ N G
5
M _ L _ S ) .
O N
D _ Y
4 ,
H _
S T _ R T _ D
W _ T H
2 5
M _ L _ S
( C Y C L _ D
1 0
M _ L _ S ) .
O N
D _ Y
3 ,
H _
S T _ R T _ D
W _ T H
1 0 0
M _ L _ S
( C Y C L _ D
T H R _ _
Q _ _ R T _ R S ,
L _ _ V _ N G
2 5
M _ L _ S ) .
O N
D _ Y
2 ,
H _
S T _ R T _ D
W _ T H
2 0 0
M _ L _ S
( C Y C L _ D
H _ L F ,
L _ _ V _ N G
1 0 0
M _ L _ S ) .
O N
D _ Y
1 ,
H _
S T _ R T _ D
W _ T H
4 0 0
M _ L _ S
( C Y C L _ D
H _ L F ,
L _ _ V _ N G
2 0 0
M _ L _ S ) Clue
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 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 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 400 MILES. WORKING BACKWARDS, ON THE FINAL DAY TOM CYCLED THE REMAINING 5 MILES. ON DAY 5, HE STARTED WITH 15 MILES (CYCLED TWO THIRDS, LEAVING 5 MILES). ON DAY 4, HE STARTED WITH 25 MILES (CYCLED 10 MILES). ON DAY 3, HE STARTED WITH 100 MILES (CYCLED THREE QUARTERS, LEAVING 25 MILES). ON DAY 2, HE STARTED WITH 200 MILES (CYCLED HALF, LEAVING 100 MILES). ON DAY 1, HE STARTED WITH 400 MILES (CYCLED HALF, LEAVING 200 MILES)