Beam Balance Surprise – Base 2 versus Base 3

 

The problem

 

          You want to weigh a mass between 1 unit to 60 units (in exact integer) using a beam balance. You are going to design a set of weights that can be used to find out the weight of this mass. What is the minimum number of such weights and what are their values?

 

 

The Binary Way

 

       We can design a set of weights :  1, 2, 4, 8, 16, 32  (units)

       We therefore need 6 weights.

 

       The method is to place the mass on one side of the beam balance and place suitable combination of the above weights on another side of the balance.

      

       1 = 1

       2 = 2

       3 = 2 + 1

       4 = 4

       5 = 4 + 1

       6 = 4 + 2

       7 = 4 + 2 + 1

               :

               :

       59 = 32 + 16 + 8 + 2 + 1

       60 = 32 + 16 + 8 + 4

 

       In the above, the technique of binary expansion is used. For example,

               60(10) = 111100(2) = 1 x 25 + 1 x 24 + 1 x 23 + 1 x 22 + 0 x 21 + 0 x 20

                       = 32 + 16 + 8 + 4 + 0 + 0

 

 

 

Unexpected method

 

       However, we can put the weights on both sides of the pans of the beam balance. For example, suppose we have two weights of 1 and 3 units, we can find out a mass of 2 units. This can be done by putting the mass (2 units) and the weight of 1 unit on one side and 3 units weight on the other side. We can therefore express :  2 = 3 – 1

 

       Then,

       We can design a set of weights :  1, 3, 9, 27, 81  (units)

       We therefore need only 5 weights!

 

(All subtraction below means that the weight is put on the other side of the balance)

       1 = 1

       2 = 3 – 1

       3 = 3

       4 = 3 + 1

       5 = 9 – 3 – 1

       6 = 9 – 3

       7 = 9 – 3 + 1

       8 = 9 – 1

       9 = 9

       10 = 9 + 1

               :

               :

       59 = 81 – 27 + 9 – 3 – 1

       60 = 81 – 27 + 9 – 3

 

 

Rationale

 

       The reason is not difficult. Remember that 1, 2, ….., 60 can be expressed as combination of

1, 2, 4, 8, 16, 32 (The Binary Numbers)

 

       But we can also write the numbers in combination in Base 3, that is 30, 31, 32, 33, 34

(or 1, 3, 9, 27, 81)

 

       For example,

              

               59(10)      = 2 x 27 + 0 x 9 + 1 x 3 + 2 x 3

                               = 2 x 33 + 0 x 32 + 1 x 31 + 2 x 30

                               = 2012(3)

 

       However, the trouble is that we have only one of each of the weight. In the above case we don’t have 2 of the weights with 27 units and 2 if the weights with 1 unit.

 

       Studying more closely, 2 of the weights can be expressed below:

               2 x 1 = 2 = 3 – 1       \         2(3) = 11(3)  (red denotes negative)

               2 x 3 = 6 = 9 – 3                    20(3) = 110(3)

               2 x 9 = 18 = 27 – 9               200(3) = 1100(3)

               2 x 27 = 54 = 81 – 27           2000(3) = 11000(3)

 

       59(10)      = 2012(3) = 2021(3)  (red denotes negative)

                       = 2111(3)

                       = 11111(3)

                       = 81 – 27 + 9 – 3 – 1  (in base 10)

       Haha! We have got it.