Weighing Only a Single Time !

[msu_math]

This is about an interesting algorithmic problem and solution. The central point of the problem is restriction on the number of weighing why the title is. At first we give a solution of a particular case of the problem and then we include a solution of the general case. Since the general solution is not obvious, we ask you to spend couple of minutes to solve the problem yourself first. We hope this practice will make our solutions much more interesting to you. 

Problem:

There are 9 different pots containing unlimited number of identical balls of different masses. Among the 9 pots, some pots contain balls of 1.1 gram and others contain balls of mass 1.0 gram. The pots are labeled by serial numbers and we are given a balance with standard mass stones (or a spring scale). We are asked to find out the pots containing balls of mass 1.1 gram by using the balance only once i.e. we can use the balance to weigh any amount of balls only a single time.

Particular Solution:

Let us first consider a particular case of the problem. Suppose, among the 9 pots, exactly 1(one) pot contains balls of 1.1 gram and rest 8(eight) pots contain balls of 1.0 gram. Let n be the serial number of the pots. Then n = 1, 2, . . , 9. Now, make a collection of balls taking n balls from the n^th pot i.e. 1 ball from the 1st pot, 2 balls from the 2nd pot, 3 balls from the 3rd and so on. Now use the balance and standard mass stones to weigh the total mass of the collection. Then the digit to the right of the decimal point in the total mass gives the serial number of the pot containing balls of mass 1.1 gram.

For example, if the 6^th pot contains the balls of mass 1.1 gram then the total mass of the aforementioned collection is

                  1x1.0 + 2x1.0 + 3x1.0 + 4x1.0 + 5x1.0 + 6x1.1 + 7x1.0 + 8x1.0 + 9x1.0 = 45.6

Here in the number 45.6, the digit 6 to the right of the decimal point indicates the serial number of the required pot.


* The solution of the general case will be posted later.

[bipasha]

well don salauddin sir..really very interesting problem..i enjoyed it

[msu_math]

General Solution:

Let us now solve the main problem. Make a collection of balls taking nx10ⁿ balls from the n^th pot i.e. 10 balls from the 1st pot, 200 balls from the 2nd pot, 3000 balls from the 3rd, 40000 balls from the 4th and so on. Now use the balance and standard mass stones to weigh the total mass of the collection. Note that, if all of the 9 pots contain balls of 1.0 gram then the total mass of the above collection will be 9876543210.0 (say, trivial mass). When some of the pots contain balls of mass 1.1 gram then the total mass of the collection must differ from the trivial mass. Then the nonzero digits in the difference of total mass and trivial mass give the serial numbers of the pots containing balls of mass 1.1 gram.

For example, if 2nd, 4th, 7th and 8th pots contain the balls of mass 1.1 gram and others contain balls of mass 1.0 gram, then total mass of the aforesaid collection is

For  n=1,   10x1.0               =           10.0
For  n=2,   200x1.1              =         220.0
For  n=3,   3000x1.0             =       3000.0
For  n=4,   40000x1.1           =      44000.0
For  n=5,   500000x1.0         =     500000.0
For  n=6,   6000000x1.0       =    6000000.0
For  n=7,   70000000x1.1     =   77000000.0
For  n=8,   800000000x1.1   =  880000000.0
For  n=9,   9000000000x1.0 = 9000000000.0

                   Total mass  =  9963547230.0

Here the difference of total mass and trivial mass is 9963547230.0 - 9876543210.0 = 87004020.0. Therefore, the nonzero digits 2, 4, 7 and 8 in the number 87004020.0 give the serial numbers of the required pots.

[Masuma Parvin]

Very interesting application.

[shirin.ns]

very interesting........

Shirin Sultana
Lecturer (Mathematics)
Dept. of Natural Sciences
Daffodil International university

[naser.te]

Interesting really.

[smriti.te]

Thanks for shearing....