Tag Archives: Jayant Narlikar

Two versions of Zeno’s paradox

  1. The story of the hare and the tortoise is well-known. The hare lost because he took a long nap on the way. But, here is an argument that shows that even if the hare were running fast, he could not overtake the tortoise, provided that the tortoise was ahead to begin with. Suppose the hare was at a point A and the tortoise was at point B ahead of A on the race track. By the time, the hare reached B, the tortoise would have moved to C ahead of B. Again when the hare reaches C; the tortoise would have moved to D ahead of C. And, so on, ad fnfinitum. So, the tortoise will always be ahead of the hare! What is the fallacy in this argument? This is one of the paradoxes of the Greek philosopher Zeno relating to infinity.
  2. Here is another of Zeno’s paradoxes. A runner wishes to cover a distance of 1 km. Zeno uses the following argument to show that he/she will never succeed if, at each stage, he/she aims at covering half the remaining distance. For to do so, first he/she will have to cover half the remaining distance. That leaves \frac{1}{4} km, of which he/she next covers half the distance. if he/she adopts the strategy of covering half the remaining distance each time. He/she has an infinity of operations lying ahead of him/her, with the destination always lying just that little bit ahead. What is wrong with this reasoning?

Ref: Fun and Fundamentals of Mathematics, Jayant V Narlikar and Mangala Narlikar, Universities Press.

More fun later,

Nalin Pithwa

Tower of Brahma

One example of very large numbers is found in the story of Brahma’s Tower supposed to be set in a temple in the holy city of Varanasi. The tower consists of 64 discs of gold of decreasing sizes with hole in the middle of each disc. To begin with, there was a tower built by placing discs in descending order of size, one on top of another around a peg. There are two other similar pegs which were empty to begin with. The priests have to transfer the tower from one peg to another making sure that each time the disc on top is transferred on to another disc (around another peg) larger in size, until the whole tower is moved to a neighbouring peg. Estimate the time taken to do this job, assuming that one move takes one second. A great deal hinges on your solution; for once the job is completed it will be the end of the universe, till Brahma creates another!!

(Ref: Fun and Fundamentals of Mathematics by Jayant V Narlikar and Mangala Narlikar).

More later,

Nalin Pithwa

Cyclic Numbers

Consider the number 2387. We can construct new numbers from the same digits by moving them cyclically:

3872, 8723, and 7238.

in general, we can construct n cyclic numbers from a number containing n digits:

N_{1}=a_{1}a_{2}a_{3} \ldots a_{n-1}a_{n}

N_{2}=a_{2}a_{3} \ldots a_{n-1}a_{n}a_{1}

\vdots

N_{3}=a_{3}a_{4}a_{5}\ldots a_{n-1}a_{n}a_{1}a_{2}

N_{n}=a_{n}a_{1}a_{2}a_{3}\ldots a_{n-1}.

What is the sum of all the cyclic numbers of such a set \{N_{1}, N_{2}, \ldots, N_{n} \}? it is easy to compute this from the above description. When we add all the numbers above, we find that in each vertical column, the sum simply is

S = a_{1}+ a_{2} + a_{3} + \ldots + a_{n-1} + a_{n}

so that if we factor it out, the remaining factor will be 111\ldots11, with the number repeated n times. In our example, chosen above, we should get

2387+3872+8723+7238=(2+3+8+7) \times 1111 = 20 \times 1111 = 22220.

You can verify that this is indeed true.

There are several puzzles related to cyclic numbers. You are most welcome to share with us.

-Nalin Pithwa

The Josephus Problem

The Jewish historian Josephus was trapped in a cave with forty other Jews, all of them trying to evade their Roman conquerors who had captured their place Jotapat. The Jews decided to kill themselves rather than be captured by the invaders. However, Josephus and a friend of his did not wish to die. To avoid expressing dissent in the midst of all present, he suggested a sequential process of killing each a member. He arranged all 41 in a circle and starting with a given person, made the rule that every third man was to be killed. Of course, in this way, the circle would get progressively shorter, till only one person would be left, who was to commit suicide. Josephus placed himself and his friend in such positions with respect to the starting one that they were the last two left and thus escaped. The question is, where were they placed initially?

To solve this problem, number all the positions and progressively eliminate every third one till only two are left. The answer is that the positions 16 and 31 are the last to go. Thus, Josephus and his friends occupied these positions.

We can check this as follows:

The original positions are:

1,2,3,4,5,6,7,8,9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41.

After the last position 41, follows the position 1 in the circle, of course. Now, let us go around the circle once, removing every third number along the way:

1, 2, 4, 5, 7, 8, 10, 11, 13, 14, 16, 17, 19, 20, 22, 23, 25, 26, 28, 29, 31, 32, 34, 35, 37, 38, 40, 41.

Continuing to go around the circle the second time, after 40 and 41, the third number is 1, which has to go. So, after the second round, we are left with:

2, 4, 7, 8, 11, 13, 16, 17, 20, 22, 25, 26, 29, 31, 34, 35, 38, 40.

And, so we continue going round and round, and successively reducing the size of the circle:

2, 4, 8, 11, 16, 17, 22, 25, 29, 31, 35, 38.

2, 4, 11, 16, 22, 25, 31, 35.

2, 4, 16, 22, 31, 35.

4, 16, 31, 35.

16, 31.

So, this is the strategy that helped Josephus to plan his own and his friend’s survival!

Ref: Fun and Fundamentals of Mathematics by Jayant V Narlikar and Mangala Narlikar

More fun offered by Professor Narlikar to be shared later,

-Nalin Pithwa

 

 

Some Arithmetic Titbits — as shared by Prof Jayant Narlikar

(Ref: Fun and Fundamentals of Mathematics by Jayant V Narlikar and Mangala Narlikar)

We will look at some interesting games and puzzles involving numbers. They will not entail anything more complicated than the four basic operations of  arithmetic: addition, subtraction, multiplication and division. To set the ball rolling, let us consider a simple three-digit number, any number, whose first and last digits are not the same. Let us take, say,

568

Now, reverse it to get 865

Next, subtract the smaller of the two from the other:

865-568=297

Now, reverse the answer and add to it. Thus,

792+297=1089.

So, you end up with the number 1089. What is so special about it? Nothing, except that, you always get this number as the final answer, no matter how you started! 

Try again with another number, say, with 841. We have the following 4 steps as above:

(1) Reversal: 148

(2) Subtraction: 841-148=693

(3) Reversal: 396

(4) Addition: 693+396=1089.

The answer of the subtraction made at the beginning must, however, be considered as a “three digit number”. If the subtraction gives the answer as 99, it is to be considered as 099. This happens if the difference between the first and the third digits of the original number is 1.

You can use this result as  a guessing game. You can ask your friend to start with any three digit number and perform these operations, in secret, without telling you. Then, you impress him/her by telling him/her the answer!

Race to 50:

Here, is a game that apparently involves addition only, but in reality also requires you to think of a strategy to win. The rules of the game are as follows:

Two players, and B, play it with each alternately adding any number from 1 to 6, to a score, starting from zero. Whosoever adds the number that brings the score to 50 wins.

The game could proceed, for example, as follows:

\begin{array}{cccc} Round \hspace{0.1 in}No & A \hspace{0.1 in }adds & B \hspace{0.1 in}adds & Total \\ 1 & 4 & 0 & 4 \\ 2 & 0 & 5 & 9 \\ 3 & 3 & 0 & 12 \\ 4 & 0 & 6 & 18 \\ 5 & 5 &  & 23 \\ 6 & 0 & 2 & 25 \\ 7 & 6 & 0 & 31 \\ 8 & & 4 & 35 \\ 9 & 1 & & 36 \\ 10 &  & 3 & 39 \\ 11 & 6 & 0 & 45 \\ 12 & 0 & 5 & 50 \end{array}

Thus, B wins.

The above is just an example. Now, you should think of a strategy so that you will necessarily win. Is it possible to ensure victory for both A and B?

More fun later with thanks to Professor Narlikar,

Nalin Pithwa

The Game of Four 4s

1.1 The Fourth Ranked King

Perhaps, nothing can give you as much fun and experience of ordinary arithmetic operations as the game of four 4s.

(Professor and eminent scientist) Jayant Narlikar, had come across it when he was in middle school. It was introduced through the following story:

A king was proud of his vast empire, and thought that his was the top-ranking kingdom in the world. He asked the scholars in his court to verify it through a world wide search. In those (pre-internet) days of the past, people had to physically travel to obtain information. And, so  the wise men travelled to the East, to the West, to the North and the South. They returned in due course with the following tidings:

“Your Majesty! Yours is the fourth largest empire on the Earth.”

The  King was disappointed and also furious. He was about to issue commands to behead these bearers of unwelcome tidings, when his Vizier stepped in.

“Sir, it is indeed a happy circumstance that you are ranked Number Four”, he added. “For, if I may be permitted to say so, of the ten primary digits 4 is the most versatile so  far as arithmetical operations are concerned.”

“Explain yourself!”, snapped the king.

The King, whatever his other shortcomings were, was well versed in elementary arithmetic. He would not be easily fobbed off.

So, the Vizier showed him the following game, a game which seemed very simple at first, but grew progressively more involved and interesting.

See for yourself, how far you can progress in the game of four 4s.

1.2 Rules of the Game

The rules of this game are simple. You are to use the number 4, four times in the well established operations of arithmetic. And you are required to construct integers 1, 2, 3, …etc.

What are the permitted operations?

(1) You can, of course, use the four fundamental operations of addition, subtraction, multiplication and division. Thus, you can express zero, one, two and three, as:

4+4-4-4=0

\frac{4}{4} \times \frac{4}{4}=1

\frac{4}{4} + \frac{4}{4}=2

\frac{4+4+4}{4}=3

II) Next, you can use the square root sign \sqrt{}. The fact that 4 is  a perfect square, helps, of course. Thus, you can construct \sqrt {4}=2, which may prove useful. The square root sign can cover big expression too, for example, \sqrt {4+4+\frac{4}{4}}=3.

Likewise, one can raise expressions to powers, example, 4^{4}=256

III) The next important operation is of decimalization, both regular and recurring. For example, the following expressions can be useful in constructing some numbers:

\frac{4}{.4}=\frac{4}{\frac{4}{10}}=10,

\frac{4}{.\overline{4}}=\frac{4}{.444444\ldots}=\frac{4}{\frac{4}{9}}=9

IV) A very useful permitted operation is the “factorial”. In general, for any integer N, we write

N! = 1 \times 2 \times 3 \ldots \times N. Therefore,

4! = 1 \times 2 \times 3 \times 4=24.

That is, about all! In other words, using all these operations on the number 4, use four and only four times, using no other number or symbol, but using parentheses as required, how  far can you go, starting with 1?

1.3 Examples

Just to see, how these operations work, let us try out a few examples:

1.3.1 The number 13

To make 13, we can proceed in many ways. For example,

13 = \frac{4!}{\sqrt{4}}+\frac{4}{4}

or 13 = 4 + 4 + \frac{\sqrt{4}}{4}

And, of course, there are other ways. Some are more elegant than others, but as long as an expression is mathematically correct, that is all that matters. You should have no difficulty in reaching 100 or going well past it. See, for example, a way of making a number like 87 next.

1.3.2 The number 87

This may appear difficult at first !! But, practice with smaller numbers will show the way:

87=4 \times 4! - \frac{4}{.\overline{4}}

Note: 

.\overline{4} means .4 recurring decimal.

Of course, there will come a stage when you are unable to make the next number with four 4s. The larger this number is the better is your score.

1.4 In the end…

The Vizier got the King interested in this game to such an extent that the latter forgot his dismay at being ranked Number 4. And, of course, instead of being punished, the Wise Men were rewarded for their tidings.

1.5 Some problems for you!

Now that you are yourself hooked onto this game, try the following problems:

  1. Construct the following numbers using four 4s:
    (a) 516 (b) 641 (c) 3634 (d) 2187
  2. Instead of four 4s, try working with three 3s. or  five 5s, to see how much better the game was with four 4s.
  3. What is the largest number that you can construct with four 4s?

So, are you having fun now? Let me know how far you reached, 

Nalin Pithwa

Reference: http://www.amazon.in/Fun-Fundamentals-Mathematics-Narlikar/dp/8173713987/ref=sr_1_1?ie=UTF8&qid=1465735705&sr=8-1&keywords=fun+and+fundamentals+of+mathematics/