A N D R _ W
H _ V _
B _ _ N
T _ _ C H _ N G
F _ R
2 1
Y _ _ R S .
B _ Y _ N C _
H _ V _
B _ _ N
T _ _ C H _ N G
F _ R
1 5
Y _ _ R S .
C _ R T
H _ V _
B _ _ N
T _ _ C H _ N G
F _ R
7
Y _ _ R S .
L _ T
A N D R _ W
=
A ,
B _ Y _ N C _
=
B
_ N D
C _ R T
=
C ,
T H _ N :
A
+
B
=
3 6
B
+
C
=
2 2
A
+
C
=
2 8
F R _ M
_ Q _ _ T _ _ N
( 2 ) ,
W _
G _ T :
B
=
2 2
-
C
F R _ M
_ Q _ _ T _ _ N
( 3 ) ,
W _
G _ T :
A
=
2 8
-
C
S _ B S T _ T _ T _
A
_ N D
B
_ N T _
_ Q _ _ T _ _ N
( 1 ) :
( 2 8
-
C )
+
( 2 2
-
C )
=
3 6
S _ L V _ N G
F _ R
C
G _ V _ S
7 .
S _ B S T _ T _ T _
( C
=
7 )
_ N T _
_ Q _ _ T _ _ N ( 2 )
_ N D
( 3 ) Clue
THE DINNER LASTED 2 HOURS. IN 2 HOURS, THE THICK CANDLE WAS 2⁄3 OF THE ORIGINAL LENGTH AND THE THIN CANDLE WAS 1⁄3 OF THE ORIGINAL LENGTH THEY STOLE 301 DIAMONDS IN TOTAL. WE NEED A NUMBER THAT IS A MULTIPLE OF 7 THAT WILL GIVE A REMAINDER OF 1 WHEN DIVIDED BY 2, 3, 4, 5, AND 6. THE LEAST COMMON MULTIPLE OF THESE NUMBERS IS 60. SO, WE NEED A MULTIPLE OF 7 THAT IS 1 GREATER THAN A MULTIPLE OF 60. 60 + 1 = 61, NOT A MULTIPLE OF 7 60 X 2 + 1 = 121, NOT A MULTIPLE OF 7 60 X 3 + 1 = 181, NOT A MULTIPLE OF 7 60 X 4 + 1 = 241, NOT A MULTIPLE OF 7 60 X 5 + 1 = 301, A MULTIPLE OF 7 ANDREW HAVE BEEN TEACHING FOR 21 YEARS. BEYONCE HAVE BEEN TEACHING FOR 15 YEARS. CURT HAVE BEEN TEACHING FOR 7 YEARS. LET ANDREW = A, BEYONCE = B AND CURT = C, THEN: A + B = 36 B + C = 22 A + C = 28 FROM EQUATION (2), WE GET: B = 22 - C FROM EQUATION (3), WE GET: A = 28 - C SUBSTITUTE A AND B INTO EQUATION (1): (28 - C) + (22 - C) = 36 SOLVING FOR C GIVES 7. SUBSTITUTE (C = 7) INTO EQUATION(2) AND (3) SPLIT THE BALLS INTO 3 GROUPS WITH 3 BALLS EACH. PICK 2 GROUPS AND USE SCALE TO DETERMINE WHICH GROUP CONTAINS THE HEAVY BALL. ONCE THE GROUP IS DETERMINED, PICK 2 BALLS AND USE THE SCALE. IF THEY ARE EQUAL, THE OTHER BALL IS THE HEAVY ONE