If both
sides of the balance
can be used, then five weights
from the powers of 3 will suffice
(1, 3, 9, 27, 81)
to measure all the weights
up to 121 grams:
1, 31, 3, 3+1, 9(3+1), 93, 9+13,
91, 9, 9+1, 9+31, 9+3, 9+3+1,
27(9+3+1), etc. However, for measuring
weights up to 60 grams there are
also many other alternative solutions,
e.g.: 1, 3, 9, 20, 27,
or even 1, 2, 7, 19, 31...
etc.
Why
it isn't possible to determine
weights from 1 to 60 using
only four weights (calibration
masses)?
Because:
A) If weights are on one side only,
then each weight can have 1
of 2 states:
1) 'on the scale' or 2) 'off'.
With only five weights, only 32
cases can be achieved (including
'all off' = 0). Therefore, 6 weights
at least are required for measuring
weights up to 60 grams.
B)
If weights are on both sides,
each weight can have 1 of 3 states:
1) 'on left side', 2) 'on right
side', or 3) 'off'.
With only four weights, 3^{4} or
81 states can be achieved. One
is 'all off' = 0. The other 80
comprise 40 pairs where the positions
of the weights are mirrored. Thus,
four weights can give only 40 unique
overall weights...
But
there is a logical way to
determine weights from 1 to
60 using only four weights!
When
considering weights from 1 
60 grams, to measure 60 discreet
weights, you need only to verify
30 discreet weights. From 1 
100 grams, you would need 50
not 100 etc. The logic is as
follows:

NOT 0, NOT 2, then 1
 NOT 2, NOT 4, then 3
 NOT 4, NOT 6, then 5 ... etc.
As
you can see you need only to verify
even weights.
For 160 grams you need only four
weights (calibration masses). If both
sides of the balance can
be used, you can use the weights
2 grams, 6 grams, 18 grams, and
54 grams as follows:
2, 62, 6, 6+2, 18(6+2), 186,
(18+2)6, 182, 18, 18+2, (18+6)2,
18+6, 18+6+2, 54(18+6+2), 54(18+6),
... , 54+6.
The
winners of the puzzle of the
month are: Gord
Steadman and Larry
Bickford. Congratulations!
