The Twelve Coin Balance Problem Answer
Here is the full solution to the classic old puzzle!
Click here to see the puzzle
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KEY TO DIAGRAMS:
? = a coin that may be too light or too heavy
H = a coin that may be too heavy
L = a coin that may be too light
G = a coin that is definitely good
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The Solution:
WEIGHING 1
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? ? ? ?
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? ? ? ?
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Weigh any four of the coins against any other four of the coins.
There are two possible outcomes:
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OUTCOME A:
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La Lb Lc Ld
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Ha Hb Hc Hd
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If they do NOT balance, then you know that the unused four coins are all good.
Also: you have two other groups of four coins. One of the groups MAY contain a light
coin (La Lb Lc Ld), OR the other group contains a heavy coin (Ha Hb Hc Hd).
Go to 2:a
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OUTCOME B:
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G G G G
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G G G G
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If they DO balance, you know all eight coins on the scales are good, and
one of the other untried four coins is a fake. You do not know if the fake is light or heavy.
Go to 2:b
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WEIGHING 2:a
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La Lb Lc Hd
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Ld G G G
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Swap one of the possibly light coins (Ld) over with one of the possibly heavy coins (Hd).
Replace the rest of the possibly heavy coins with three of the good coins.
There are three possible
outcomes:
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OUTCOME A:
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G G G G
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G G G G
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If the two sides balance then we know all the coins used are good.
Therefore the fake is Ha Hb or Hc and we also know the fake
is heavy.
Go to 3:a
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OUTCOME B:
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La Lb Lc G
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G G G G
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If La Lb Lc Hd is lighter then we know that swapping Hd and Ld over made no difference,
and therefore they
both must be good. Therefore the fake is La Lb or Lc and we also know the fake is
light.
Go to 3:b
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OUTCOME C:
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G G G Hd
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Ld G G G
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If La Lb Lc Hd is now heavier, none of La, Lb or Lc can be too light so they are good.
Therefore either Hd is too heavy OR Ld is too light.
Go to 3:c
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Weigh three of the untried coins against three good coins.
There are three possible
outcomes:
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Just out of interest...
Here's an old silver three penny piece and also
a six penny piece. (The six penny piece is about 2cm across which is roughly the
size of a modern penny.) It's sad to think that even the old six pence
is only worth 2½p.
Heads
Tails
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OUTCOME A:
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Ha Hb Hc
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G G G
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If the untried coins are heavier than the three good coins, then we know that
the fake is Ha Hb or Hc and we also know the fake is
heavy.
Go to 3:a
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OUTCOME B:
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La Lb Lc
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G G G
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If the untried coins are lighter than three good coins, then we know that
the fake is La Lb or Lc and we also know the fake is
light.
Go to 3:b
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If the two sides balance then we know all the coins used are good.
Therefore the last untried coin is the fake, but we don't know if it's
light or heavy.
Go to 3:d
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We know one of Ha, Hb or Hc is heavy. Weigh Ha against Hb. There are
three possible outcomes:
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Ha is heavier and so is the heavy fake.
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Hb is heavier and so is the heavy fake.
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Ha and Hb balance so are both good. Therefore Hc is the heavy fake.
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We know one of La, Lb or Lc is light. Weigh La against Lb. There are
three possible outcomes:
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La is lighter and so is the light fake.
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Lb is lighter and so is the light fake.
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La and Lb balance so are both good. Therefore Lc is the light fake.
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We know either Ld is light or Hd is heavy. Weigh Ld against a good coin.
There are two possible outcomes:
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Did you spot our fake?
Here's a real old threepenny piece!
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Fake! |
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Ld is lighter than the good coin and so is the light fake.
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Ld weighs the same as the good coin so is also good. Therefore Hd is the heavy fake.
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We know the last untried coin is either light or heavy, so weigh
it against a good coin. There are two possible outcomes:
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The fake coin is light.
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The fake coin is heavy.
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This is clever stuff - so if you've understood it WELL DONE!
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Murderous Maths Main Index Page
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This solution to the Twelve Coin Balance Problem is copyright
© Kjartan Poskitt 2004
and may not be reproduced in any form for commercial gain without permission. Thank you.
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