rec.puzzles Archive

You have 5 sacks of gold and a normal scale.

logic/weighing/five.sacks New, unsolved

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.  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?

[ Editor: the scale is a "normal" scale that reads the weight of what is
put on it, not a balance; compare logic/weighing/balance. ]
Show solution
No solution was found in the collected postings.

[ Editor: the following attempt was posted, and refuted. ]

Weigh A+B.  They weigh n oz.

Weigh C+D.
   If they also weigh n oz., you know that E is the guilty sack, and you
can easily establish in which direction.
   If they weigh m (>n) oz., then

Weigh A+C.
   If they weigh n oz., then D is heavy.
   If they weigh m oz., then A is heavy.

If m<n, weigh A+C anyway - you can work out the results.

-- Nick Wedd <nick@maproom.demon.co.uk>

Doesn't work.  Suppose A weighs 1, and the rest weigh 2.  C+D=m > A+B=n.
A+C=n, but D isn't heavy as you suggest.

-- Lucky

[ Editor: likewise, if B is light, A+C weighs m, but A is not heavy. ]

[ Editor: further material on this puzzle, edited separately from other collected postings (batch 13); it may repeat some of the above. ]

[ Editor: problem as stated there: ]
You have 5 sacks of gold and a normal scale (not a pan balance, but a
scale like a bathroom scale, on which you put an object or objects and
get a total weight).  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> <aacmrmaze@aol.com>

ObPuzzle: Prove n weighings suffice for 2^(n-1) + 2^(n-2) - 1 sacks,
but not for 2^(n-1) + 2^(n-2) sacks.

-- Lucky


No solution was found in the collected postings.

[ Editor: the only posting collected quotes the problem with "<a
solution snipped>" and adds the ObPuzzle. ]