You have 5 sacks of gold and a normal scale (one that gives a total weight, like a bathroom scale, not a pan balance).

logic/weighing/scale.five.sacks New

New entry, edited from material collected for the archive and not yet in a released version.

Problem

You have 5 sacks of gold and a normal scale (one that gives a total
weight, like a bathroom scale, not a pan balance).  You know one of the
sacks weighs a different amount than the other sacks, but you don't
know if it is heavier or lighter.  How can you find the counterfeit
sack, and whether it weighs more or less than the others, in only three
weighings?

-- Brian, via John <www.MAZES.com>

[ Editor: lightly condensed from Brian's clarified question; the
  parenthesis paraphrases his explanation of "normal scale". ]

Show solution

Call the sacks A, B, C, D, and E.  Weigh A+B+C.  Weigh C+D+E.  If
A+B+C = C+D+E, weigh E to determine whether it weighs more or less than
1/3 of A+B+C (or C+D+E).

If A+B+C > C+D+E, then we know that either A or B is heavy or D or E
is light (4 possibilities).  Weighing A+D will produce a different
result for each of these possibilities, the 4 different results
corresponding to whether A+D = (A+B+C)-C = A+B or
A+D = (C+D+E)-C = D+E and to whether C equals (A+B+C)/3 or (C+D+E)/3.

  A+D = A+B+C - (A+B+C)/3: E is light, b/c the only way that A+D can
                           be 2/3 of A+B+C is if E is the counterfeit.
  A+D = C+D+E - (C+D+E)/3: B is heavy, b/c the only way that A+D can
                           be 2/3 of C+D+E is if B is the counterfeit.
  A+D = A+B+C - (C+D+E)/3: A is heavy, C=(C+D+E)/3 means C=D=E and
                           b/c A+D = A+B+C - C = A+B means B=D(=C=E),
                           so A is the counterfeit.
  A+D = C+D+E - (A+B+C)/3: D is light, for reasons similar to above.

-- Lucky

[ Editor: in the first case (A+B+C = C+D+E) the counterfeit is C, and
  E is a genuine sack, so E against (A+B+C)/3 tells whether C is heavy
  or light.  The case A+B+C < C+D+E is symmetric. ]