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Messages - msu_math

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76
History of Mathematics / Re: History of Mathematics
« on: November 12, 2012, 12:41:10 PM »
Thanks a lot sir. This history covers almost all step by step evolution of Mathematics. At present, a group of Mathematicians in Max-Planck institute is trying to give a new face in mathematical studies which will bring a rapid change Mathematics in near future. I am preparing a separate article on this topic and will post very soon.   

77
Interesting Maths / A Necklace with 100 Beads
« on: November 09, 2012, 03:25:20 PM »
Problem:

There are 100 beads on a necklace. 1 of them is red, the rest are blue. Working in the clockwise direction, we start from the red bead and remove every other bead. Right when the red bead is removed, how many blue beads are there left on the necklace?

Note: The first bead removed is the blue bead that is right next to the red bead in the clockwise direction. Removing every other bead means that we skip a bead, remove a bead. Skip a bead, remove a bead, etc.


Solution:     

Since initially we have 100 (an even number), on the first cycle we remove 100/2 = 50 blue balls leaving 50 balls with one of them red. The first cycle stopped (the last bead removed in the first cycle) is the blue ball on the left of the red ball. Now the second cycle is like the first only we have now 50 balls and we also begin removing the blue ball right next to the red ball (since we stopped removing the blue ball on the left of the red in the first cycle). Hence we remove 50/2 = 25 blue balls leaving 25 balls with one of them red. Now we have 25 balls and it can be shown that on the third cycle, we finally be able to remove the red bead since we only have an odd number of balls and we also begin removing the bead on the right of red. Hence with the red ball marked as 25 and the ball right next to it as 1, 2,.. ,24, on the third cycle we remove beads 1, 3, 5, …, 25 or a total of 13 beads leaving as with 25 – 13 = 12 blue beads.

Answer: 12 beads.

78
Interesting Maths / Weighing Only a Single Time !
« on: November 09, 2012, 03:18:50 PM »
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 nth 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 6th 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.

79
This is about an interesting algorithmic problem and solution. 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 nth 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 6th 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.

General Solution:

Let us now solve the main problem. Make a collection of balls taking nx10n balls from the nth 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.

80
Science Discussion Forum / Re: 101 Simple Truths We Often Forget
« on: January 24, 2012, 06:48:53 PM »
Great !

81
Congratulation to Mr. Mijanur Rahman. May this book fulfills the demand of the readers.

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