T H _
T _ T _ L
W _ _ G H T
_ F
T H _
D _ G ,
C _ T
_ N D
R _ B B _ T
_ S
2 7 K G .
T H _
D _ G ,
C _ T
_ N D
R _ B B _ T
W _ _ G H
1 7 K G ,
7 K G
_ N D
3 K G
R _ S P _ C T _ V _ L Y .
L _ T
D = D _ G ,
C = C _ T ,
R = R _ B B _ T .
W _
C _ N
C _ M _
W _ T H
T H _
F _ L L _ W _ N G
_ Q _ _ T _ _ N S :
C
+
R
=
1 0
D
+
R
=
2 0
D
+
C
=
2 4
F R _ M
1 S T
_ Q _ _ T _ _ N ,
W _
G _ T :
R
=
1 0
- C
S _ B S T _ T _ T _
T H _ T
_ N T _
2 N D
_ Q _ _ T _ _ N
_ N D
W _
G _ T :
D
+
( 1 0
-
C )
=
2 0
S _ L V _ N G
F _ R
D
G _ V _ S :
D
=
C
+
1 0
S _ B S T _ T _ T _
T H _ T
_ N T _
3 R D
_ Q _ _ T _ _ N
_ N D
W _
G _ T :
( C
+
1 0 )
+
C
=
2 4
S _ L V _ N G
F _ R
C
G _ V _ S
7 .
W _
C _ N
T H _ N
S _ B S T _ T _ T _
C = 7
_ N
T H _
_ T H _ R
_ Q _ _ T _ _ N S
T _
G _ T
T H _
W _ _ G H T
_ F
D
_ N D
R Clue
THE MINIMUM NUMBER OF WEIGHTS REQUIRED IS FIVE AND THESE SHOULD WEIGHT 1, 3, 9, 27 AND 81 POUNDS. THE MERCHANT HAS TO USE A BALANCE WEIGHING SCALE TO DO THE JOB. TO WEIGH 2 POUNDS, HE'LL HAVE TO PUT THE 3 POUND WEIGHT ON ONE PAN AND 1 POUND WEIGHT ON THE OTHER PAN. TO WEIGH 5 POUNDS, HE'LL HAVE TO PUT THE 9 POUND WEIGHT ON ONE PAN AND 1 AND 3 POUND WEIGHTS ON THE OTHER PAN THERE ARE 9 SHEEP AND 18 CHICKENS. LET S BE THE NUMBER OF SHEEP AND C BE THE NUMBER OF CHICKENS. SO: 2S = C 5S + 3C = 99 WE CAN REPHRASE THE FIRST EQUATION, SO: 6S - 3C = 0 AND THEN WE CAN ADD THIS TO THE SECOND EQUATION, WHICH YIELDS: 11S = 99 BY SOLVING FOR S, WE FIND THAT S EQUALS 9. BY SUBSTITUTING BACK IN ONE OF THE ORIGINAL EQUATIONS, WE FIND THAT C EQUALS 18. SO THERE ARE NINE SHEEP AND EIGHTEEN CHICKENS THE TOTAL WEIGHT OF THE DOG, CAT AND RABBIT IS 27KG. THE DOG, CAT AND RABBIT WEIGH 17KG, 7KG AND 3KG RESPECTIVELY. LET D=DOG, C=CAT, R=RABBIT. WE CAN COME WITH THE FOLLOWING EQUATIONS: C + R = 10 D + R = 20 D + C = 24 FROM 1ST EQUATION, WE GET: R = 10 -C SUBSTITUTE THAT INTO 2ND EQUATION AND WE GET: D + (10 - C) = 20 SOLVING FOR D GIVES: D = C + 10 SUBSTITUTE THAT INTO 3RD EQUATION AND WE GET: (C + 10) + C = 24 SOLVING FOR C GIVES 7. WE CAN THEN SUBSTITUTE C=7 IN THE OTHER EQUATIONS TO GET THE WEIGHT OF D AND R 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