T H _ R _
_ R _
2
M _ N ,
5
W _ M _ N
_ N D
1 3
C H _ L D R _ N .
L _ T
T H _
N _ M B _ R
_ F
M _ N ,
W _ M _ N
_ N D
C H _ L D R _ N
B _
M ,
W ,
_ N D
C
R _ S P _ C T _ V _ L Y .
T H _ N
T H _
P R _ B L _ M
S T _ T _ S
T H _ T :
M
+
W
+
C
=
2 0
3 M
+
1 . 5 W
+
0 . 5 C
=
2 0
M _ L T _ P L Y _ N G
_ Q _ _ T _ _ N
2
B Y
2
G _ V _ S :
6 M
+
3 W
+
C
=
4 0
S _ B T R _ C T _ N G
T H _
_ B _ V _
_ Q _ _ T _ _ N
F R _ M
T H _
F _ R S T
_ Q _ _ T _ _ N
G _ V _ S :
5 M
+
2 W
=
2 0
T H _
_ N _ Q _ _
S _ L _ T _ _ N
_ S
M
=
2 ,
W
=
5
_ N D
C
=
1 3 Clue
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 THE PROBABILITY IS 1/2. THERE ARE INITIALLY 4 POSSIBILITIES: GIRL-GIRL, GIRL-BOY, BOY-GIRL AND BOY-BOY. SINCE THE OLDER CHILD IS A BOY, WE CAN RULE OUT THE GIRL-GIRL AND GIRL-BOY COMBINATIONS A HALF DOLLAR, A QUARTER, AND FOUR DIMES. YOU OBVIOUSLY CAN'T HAVE ANY PENNIES, BECAUSE YOU COULDN'T HAVE 1-4 PENNIES, SINCE $1.15 IS A MULTIPLE OF 5, AND IF YOU HAD 5 PENNIES OR MORE YOU COULD CHANGE A NICKEL. YOU CAN'T HAVE MORE THAN 1 NICKEL, OTHERWISE YOU COULD CHANGE A DIME. AND IF YOU HAD 1 NICKEL YOU COULDN'T HAVE BUT 1 DIME, OTHERWISE YOU COULD CHANGE A QUARTER. THEN IF YOU HAD 1 NICKEL AND 1 DIME YOU'D HAVE 15 CENTS. THAT WOULD LEAVE $1.00 AND YOU COULDN'T MAKE THAT REMAINING $1.00 WITH QUARTERS AND HALVES WITHOUT EITHER BEING ABLE TO CHANGE A HALF OR A DOLLAR. SO YOU CAN'T HAVE ANY PENNIES OR NICKELS. SINCE YOU CAN'T HAVE ANY PENNIES OR NICKELS, YOU CAN'T POSSIBLY MAKE $1.15 WITH DIMES AND HALVES ONLY, SO YOU MUST HAVE AT LEAST 1 QUARTER. YOU CAN'T HAVE MORE THAT ONE QUARTER, OTHERWISE YOU COULD CHANGE A HALF. SO YOU MUST HAVE EXACTLY 1 QUARTER. THAT LEAVES 90 CENTS, TO MAKE WITH A HALF DOLLAR AND DIMES. SO THAT'S 1 HALF AND 4 DIMES. AND WITH ONLY 4 DIMES YOU CAN'T CHANGE A HALF DOLLAR THERE ARE 2 MEN, 5 WOMEN AND 13 CHILDREN. LET THE NUMBER OF MEN, WOMEN AND CHILDREN BE M, W, AND C RESPECTIVELY. THEN THE PROBLEM STATES THAT: M + W + C = 20 3M + 1.5W + 0.5C = 20 MULTIPLYING EQUATION 2 BY 2 GIVES: 6M + 3W + C = 40 SUBTRACTING THE ABOVE EQUATION FROM THE FIRST EQUATION GIVES: 5M + 2W = 20 THE UNIQUE SOLUTION IS M = 2, W = 5 AND C = 13