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 Five Weights and a Broken Scale (Posted on 2004-08-02)
You have to sort five weights weighing 51g, 52g, 53g, 54g, and 55g. You have a balance scale with which you can compare the weights. But after solving so many sorting puzzles, it is starting to break down.

If the difference between two weights is greater than 1.5g, the scale will correctly determine which side is heavier. If the difference between the weights is less than 1.5g or equal, the scale will indicate the weights are equal.

Sort the weights in the smallest number of weighings.

 See The Solution Submitted by Brian Smith Rating: 3.5556 (9 votes)

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 re: The rest of my comment | Comment 5 of 17 |
(In reply to The rest of my comment by Charlie)

An example of how the number of weighings could be reduced in some instances is the following:

Say in weighing a+b vs c+d and a+c vs b+e, in both instances the latter pan was heavier, so you use the top segment of the table, which indicates to weigh a vs b, a vs c, b vs c and b+d vs c+e.  If the weighing b+d vs c+e is done first, then:

If c+d is the heavier, you don't need to weigh a vs c, just a vs b and b vs c. And in fact, if you get an unbalanced indication on the first of these two weighings, you don't need to proceed further.

If c+d balances b+d, however, you'd need weigh only a vs b: if the latter were heavier, the order would be acbde, otherwise abced, after just 4 weighings in this instance.

And if b+d were the heavier, you'd need weigh only a vs c and b vs c.

The instance of the initial two weighings being unbalanced in opposite directions is one in which such shortcuts will not work.  I'm fairly sure 6 is the minimum number of weighings that will assure an ordering in all circumstances.

 Posted by Charlie on 2004-08-03 11:41:38

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