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 Unreliable scale (Posted on 2008-02-28)
You are given nine coins. The coins are identical in all respects except that one is slightly heavier than the remaining eight. You also have a somewhat unreliable comparison (or, balance) scale. The scale may give up to one false reading, i.e., the scale may work perfectly every time, or it may produce just one erroneous result. The only possible erroneous results are either (1) indicating an inequality in either direction when, in fact, the two sides are balanced, or (2) indicating equality when the right side is heavier.

Define a procedure for identifying the heavy coin with three weighings. The procedure must also determine which, if any, of the weighings were erroneous.

 See The Solution Submitted by FrankM Rating: 3.0000 (3 votes)

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 Solution | Comment 2 of 3 |

This is a good puzzle.  At first it seems three weighings are insufficient, but the fact that the scale will give at most one false result adds enough to get a solution.

Label the coins A-I.  Make the first two weighings ABC/DEF and ADG/BEH.  After the two weighings there are three possible heavy coins, outlined below.

ABC/DEF|ADG/BEH|both | #1  | #2
(GHI) | (CFI) |right|wrong|wrong
-------+-------+-----+-----+-----
>   |   >   |  A  |  G  |  C
-------+-------+-----+-----+-----
>   |   =   |  C  |  I  |  B
-------+-------+-----+-----+-----
>   |   <   |  B  |  H  |  C
-------+-------+-----+-----+-----
=   |   >   |  G  |  D  |  I
-------+-------+-----+-----+-----
=   |   =   |  I  |  F  |  H
-------+-------+-----+-----+-----
=   |   <   |  H  |  E  |  I
-------+-------+-----+-----+-----
<   |   >   |  D  |  G  |  F
-------+-------+-----+-----+-----
<   |   =   |  F  |  I  |  E
-------+-------+-----+-----+-----
<   |   <   |  E  |  H  |  F
-------+-------+-----+-----+-----
|       | (X) | (Y) | (Z)

Call the three possible heavy coins X, Y and Z.  Let X be the heavy coin if both two weighings were correct.  Let Y be the heavy coin if the first weighing was wrong, and Let Z be the heavy coin if the second weighing was wrong.  Make the third weighing X/Y.  There are six possibilities as follows:

X/Y| right  | wrong
(Z)|(#wrong)|(#wrong)
---+--------+--------
> | X (0)  | Z (2)
= | Z (1)  | Y (2)
< | Y (1)  | Z (2)

Since the scale is wrong at most once, the third weighing must be accurate. X>Y does mean X is heavy (and all weighings accurate), X=Y does mean Z is heavy (and second weighing in error), and X<Y does mean Y is heavy (and first weighing in error).

 Posted by Brian Smith on 2008-03-01 01:50:21

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