You have a balance scale, with one arm twice the length of the other.

logic/weighing/uneven.arms New

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

Problem

You have a balance scale, with one arm twice the length of the other.
What is the smallest number of weights required to weigh all integer
weights from 1 to 1000?

-- Stephen Tavener <Taveners@LILHD.Logica.Com>

Show solution

10 weights suffice, e.g. 2,4,8,16,32,64,128,256,512,1024 works
up to 1024 and 1,2,2.5,5,10,20,40,80,160,320 works up to 1280.

-- Chris Long <clong@remus.rutgers.edu>

Note that we now have an upper limit of 10.
A lower limit is 7, since each weight can be 1) on the short side of the
scales 2) on the long side, or 3) not used.  This means that for n
weights we have 3**n possible combinations, and 3**6 < 1000 < 3**7.

Some results from computer search:

#of weights        Range                Weights
1                1                2
2                1-4                1,3
3                1-12                2,4,18
4                1-30                2,4,8,46
5                1-74                1,5,12,16,20 **

**My programme went haywire at this point

=> increasing by about 2.5 for 3->4, 4->5.  If this holds for higher
weights, then I'd expect 8 weights to bust the 1000 barrier.

-- Stephen Tavener <Taveners@LILHD.Logica.Com>

[ Editor: no later result was found in the collected postings. ]