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, 3-1, 3, 3+1, 9-(3+1), 9-3, 9+1-3,
9-1, 9, 9+1, 9+3-1, 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, 34 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 1-60 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, 6-2, 6, 6+2, 18-(6+2), 18-6,
(18+2)-6, 18-2, 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!
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