S P L _ T
T H _
C _ _ N S
_ N T _
3
G R _ _ P S
-
2
G R _ _ P S
W _ T H
3
C _ _ N S
_ _ C H
_ N D
1
G R _ _ P
W _ T H
2
C _ _ N S .
N _ W
P _ T
T H _
2
G R _ _ P S
_ F
3
C _ _ N S
_ N
T H _
S C _ L _ .
S C _ N _ R _ _
# 1 :
B _ T H
G R _ _ P S
W _ _ G H
T H _
S _ M _ .
T H _ S
M _ _ N S
T H _ T
T H _
C _ _ N T _ R F _ _ T
C _ _ N
_ S
_ N
T H _
G R _ _ P
W _ T H
2
C _ _ N S .
S _
T _ K _
T H _
2
C _ _ N S
F R _ M
T H _ T
G R _ _ P
_ N D
_ S _
T H _
S C _ L _
T _
D _ T _ R M _ N _
T H _
C _ _ N T _ R F _ _ T
C _ _ N .
S C _ N _ R _ _
# 2 :
O N _
G R _ _ P
W _ _ G H S
L _ S S
T H _ N
T H _
_ T H _ R .
T _ K _
_ N Y
2
C _ _ N S
F R _ M
T H _ S
G R _ _ P
_ F
3
C _ _ N S .
I F
T H _ Y
W _ _ G H
T H _
S _ M _ ,
T H _ N
T H _
3 R D
C _ _ N
_ S
T H _
C _ _ N T _ R F _ _ T
C _ _ N .
I F
T H _ Y
D _ N ' T
W _ _ G H
T H _
S _ M _ ,
T H _ N
T H _
L _ G H T _ R
_ N _
_ S
_ B V _ _ _ S L Y
T H _
C _ _ N T _ R F _ _ T
C _ _ N Clue
8 WEIGHINGS ARE REQUIRED TO FIND OUT THE HEAVY BALL. DIVIDE THE BALLS INTO 3 GROUPS OF 2187 BALLS EACH. PUT 2 GROUPS ON THE SCALE AND DETERMINE WHICH GROUP IS HEAVIER. IF BOTH GROUPS ARE EQUAL IN WEIGHT, THE HEAVIER BALL IS IN THE 3RD GROUP. REPEAT THE PROCESS BY BREAKING THE GROUP WITH THE HEAVIER BALL INTO 3 SMALLER GROUPS OF BALLS AGAIN. FOR THE 2ND ROUND, EACH GROUP WILL HAVE 729 (2187 / 3 ) BALLS EACH. THIS PROCESS HAS TO BE REPEATED 8 TIMES. 1ST WEIGHING - 6561 BALLS ARE DIVIDED INTO 3 GROUPS OF 2187 BALLS EACH. 2ND WEIGHING - 2187 BALLS ARE DIVIDED INTO 3 GROUPS OF 729 BALLS EACH. 3RD WEIGHING - 729 BALLS ARE DIVIDED INTO 3 GROUPS OF 243 BALLS EACH. 4TH WEIGHING - 243 BALLS ARE DIVIDED INTO 3 GROUPS OF 81 BALLS EACH. 5TH WEIGHING - 81 BALLS ARE DIVIDED INTO 3 GROUPS OF 27 BALLS EACH. 6TH WEIGHING - 29 BALLS ARE DIVIDED INTO 3 GROUPS OF 9 BALLS EACH. 7TH WEIGHING - 9 BALLS ARE DIVIDED INTO 3 GROUPS OF 3 BALLS EACH. 8TH WEIGHING - 3 BALLS ARE DIVIDED INTO 3 GROUPS OF 1 BALL EACH TAKE ONE COIN FROM THE FIRST BAG, TWO FROM THE SECOND, THREE FROM THE THIRD, AND SO ON, UNTIL YOU HAVE TEN COINS FROM THE TENTH. THEN STACK AND WEIGH THE 55 COINS (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10). SINCE EACH GOLD COIN WEIGH TEN GRAMS, IF ALL HAD BEEN GOLD, THE SCALE WOULD HAVE READ 550 GRAMS. THE AMOUNT BY WHICH THE WEIGHT WAS TOO LIGHT INDICATED THE NUMBER OF SILVER COINS AND THE NUMBER OF THE THE SILVER BAG. FOR INSTANCE, IF THE WEIGHT WAS 543 GRAMS, IT WOULD INDICATE THAT 7 SILVER COINS (550 - 543 = 7) HAD BEEN WEIGHED WITH THE GOLD COINS AND THAT THE REST OF THE SILVER COINS WERE IN THE SEVENTH BAG PULL ONE COIN OUT OF THE CHEST MARKED "50 GOLD AND 50 SILVER COINS". IF IT'S A GOLD COIN, YOU KNOW THAT IT'S THE CHEST WITH 100 GOLD COINS. THE CHEST LABELLED "100 GOLD COINS" WILL THEN HAVE 100 SILVER COINS AND THE CHEST LABELLED "100 SILVER COINS" WILL HAVE 50 GOLD AND 50 SILVER COINS. IF THE FIRST COIN THAT YOU PULLED OUT FROM THE CHEST LABELLED "50 GOLD AND 50 SILVER COINS" IS A SILVER COIN, THEN YOU SOLVE THE PROBLEM IN THE SAME GENERAL WAY SPLIT THE COINS INTO 3 GROUPS - 2 GROUPS WITH 3 COINS EACH AND 1 GROUP WITH 2 COINS. NOW PUT THE 2 GROUPS OF 3 COINS ON THE SCALE. SCENARIO #1: BOTH GROUPS WEIGH THE SAME. THIS MEANS THAT THE COUNTERFEIT COIN IS IN THE GROUP WITH 2 COINS. SO TAKE THE 2 COINS FROM THAT GROUP AND USE THE SCALE TO DETERMINE THE COUNTERFEIT COIN. SCENARIO #2: ONE GROUP WEIGHS LESS THAN THE OTHER. TAKE ANY 2 COINS FROM THIS GROUP OF 3 COINS. IF THEY WEIGH THE SAME, THEN THE 3RD COIN IS THE COUNTERFEIT COIN. IF THEY DON'T WEIGH THE SAME, THEN THE LIGHTER ONE IS OBVIOUSLY THE COUNTERFEIT COIN