Monthly Archives: May 2017

Something Cute I Never Noticed Before About Infinite Sums

Source: Something Cute I Never Noticed Before About Infinite Sums

Arithmetic Puzzle

Gaurish4Math

Following is a very common arithmetic puzzle that you may have encountered as a child:

Express any whole number $latex n&bg=ffffff$ using the number 2 precisely four times and using only well-known mathematical symbols.

This puzzle has been discussed on pp. 172 of Graham Farmelo’s “The Strangest Man“, and how Paul Dirac solved it by using his knowledge of “well-known mathematical symbols”:

$latex displaystyle{n = -log_{2}left(log_{2}left(2^{2^{-n}}right)right) = -log_{2}left(log_{2}left(underbrace{sqrt{sqrt{ldotssqrt{2}}}}_text{n times}right)right)}&bg=ffffff$

This is an example of thinking out of the box, enabling you to write any number using only three/four 2s. Though, using a transcendental function to solve an elementary problem may appear like an overkill.  But, building upon such ideas we can try to tackle the general problem, like the “four fours puzzle“.

This post on Puzzling.SE describes usage of following formula consisting of  trigonometric operation $latex cos(arctan(x)) = frac{1}{sqrt{1+x^2}}&bg=ffffff$ and $latex tan(arcsin(x))=frac{x}{sqrt{1-x^2}}&bg=ffffff$ to obtain the square…

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APJ Abdul Kalam : How to manage success and failure

It is examinations &/or results of examinations time in India. Especially, the ruthlessly, mercilessly competitive, intense IITJEE Advanced…

So, here is some advice for a life time from our revered, adored APJ Abdul Kalam. Actually, this advice applies to all adult stages in life…

Three in a row !!!

If my first were a 4,

And, my second were a 3,

What I am would be double,

The number you’d see.

For I’m only three digits,

Just three in a row,

So what must I be?

Don’t say you don’t know!

Cheers,

Nalin Pithwa.

Follow Descartes’ Historically Famous Problems !

Problem 1:

Three circles touching one another externally have radii r_{1}, r_{2} and r_{3}. Determine the radii of the two circles that can be drawn touching all the three circles.

Problem 2:

Consider a circle, say (numbered 1) of unit radius 1. Inside this circle, two circles are drawn (say, numbered 2 and 3), each of radius \frac{1}{2}, which touch each other externally and the first circle internally. Determine the radius of the fourth circle, which touches circles 2 and 3 externally and circle 1 internally. Determine the radius of the fifth circle, which touches each of the circles 2, 3, and 4 externally. Determine the radius of the sixth circle, which touches circles 2 and 4 externally and circle 1 internally. One might notice that curvature of all such circles drawn within the first circle has integer curvature!

It is such historically famous problems (within scope of IITJEE Mains and IITJEE Advanced Maths) which all students should try to internalize all the concepts of Math for IITJEE. Also, in a similar vein, you should practice deriving all basic formulae, relationships of co-ordinate geometry.

More later,

Nalin Pithwa.

Solutions to Birthday Problems: IITJEE Advanced Mathematics

In the following problems, each year is assumed to be consisting of 365 days (no leap year):

Problem 1:

What is the least number of people in a room such that it is more likely than not that at least two people will share the same birthday?

Solution 1:

The probability of the second person having a different birthday from the first person is \frac{364}{365}. The probability of the first three persons having different birthdays is \frac{364}{365} \times \frac{363}{365}. In this way, the probability of all n persons in a room having different birthdays is P(n) = \frac{364}{365} \times \frac{363}{365} \times \frac{362}{365} \times \ldots \frac{365-n+1}{365}. For the value of n, when P(n) falls just below 1/2 is the least number of people in a room when the probability of at least two people having the same birthday becomes greater than one half (that is, more likely than not). Now, one can make the following table:

\begin{tabular}{|c|c|}\hline    N & P(n) \\ \hline    2 & 364/365 \\ \hline    3 & 0.9918 \\ \hline    4 & 0.9836 \\ \hline    5 & 0.9729 \\ \hline    6 & 0.9595 \\ \hline    7 & 0.9438 \\ \hline    8 & 0.9257 \\ \hline    9 & 0.9054 \\ \hline    10 & 0.8830 \\ \hline    11 & 0.8589 \\ \hline    12 & 0.8330 \\ \hline    13 & 0.8056 \\ \hline    14 & 0.7769 \\ \hline    15 & 0.7471 \\ \hline    16 & 0.7164 \\ \hline    17 & 0.6850 \\ \hline    18 &0.6531 \\ \hline    19 & 0.6209 \\ \hline    20 & 0.5886 \\ \hline    21 & 0.5563 \\ \hline    22 & 0.5258 \\ \hline    23 & 0.4956 \\ \hline    \end{tabular}

Thus, the answer is 23. One may say that during a football match with one referee, it is more likely than not that at least two people on the field have the same birthday! 🙂 🙂 🙂

Problem 2:

You are in a conference. What is the least number of people in the conference (besides you) such that it is more likely than not that there is at least another person having the same birthday as yours?

Solution 2:

The probability of the first person having a different birthday from yours is \frac{364}{365}. Similarly, the probability of the first two persons not having the same birthday as yours is \frac{(364)^{2}}{(365)^{2}}. Thus, the probability of n persons not  having the same birthday as yours is \frac{(364)^{n}}{(365)^{n}}. When this value falls below 0.5, then it becomes more likely than not that at least another person has the same birthday as yours. So, the least value of n is obtained from (\frac{364}{365})^{n}<\frac{1}{2}. Taking log of both sides, we solve to get n>252.65. So, the least number of people required is 253.

Problem 3:

A theatre owner announces that the first person in the queue having the same birthday as the one who has already purchased a ticket will be given a free entry. Where (which position in the queue) should one stand to maximize the chance of earning a free entry?

Solution 3:

For the nth person to earn a free entry, first (n-1) persons must have different birthdays and the nth person must have the same birthday as that of one of these previous (n-1) persons. The probability of such an event can we written as

P(n) = [\frac{364}{365} \times \frac{363}{365} \times \frac{362}{365} \times \ldots \frac{365-n+2}{365}] \times \frac{n-1}{365}

For a maximum, we need P(n) > P(n+1). Alternatively, \frac{P(n)}{P(n+1)} >1. Using this expression for P(n), we get \frac{365}{365-n} \times \frac{n-1}{n} >1. Or, n^{2}-n-365>0. For positive n, this inequality is satisfied first for some n between 19 and 20. So, the best place in the queue to get a free entry is the 20th position.

More later,

Nalin Pithwa.

Three More Quickies

Reference:

“Professor Stewart’s Cabinet of Mathematical Curiosities”, Ian Stewart.

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  1. If five dogs dig five holes in five days, how long does it take ten dogs to dig ten holes? Assume that they all dig at the same rate all the time and all holes are the same size.
  2. A woman bought a parrot in a pet shop. The shop assistant, who always told the truth said, “I guarantee that this parrot will repeat every word it hears.” A week later, the woman took the parrot back, complaining that it hadn’t spoken a single word. “Did anyone talk to it?” asked the suspicious assistant. “Oh, yes.”. What is the explanation?
  3. The planet Nff-Pff in the Anathema Galaxy is inhabited by precisely two sentient beings, Nff and Pff. Nff lives on a large continent, in the middle of which is an enormous lake. Pff lives on an island in the middle of the lake. Neither Nff nor Pff can swim, fly or teleport: their only form of transport is to walk on dry land. Yet, each morning, one walks to the other’s house for breakfast. Explain.

Shared by Nalin Pithwa.

Birthday Probability Problems: IITJEE Advanced Mathematics

In the following problems, each year is assumed to be consisting of 365 days (no leap year):

  1. What is the least number of people in a room such that it is more likely than not that at least two people will share the same birthday?
  2. You are in a conference. What is the least number of people in the conference (besides you) such that it is more likely than not that there is at least another person having the same birthday as yours?
  3. A theatre owner announces that the first person in the queue having the same birthday as the one who has already purchased a ticket will be given a free entry. Where (which position in the queue) should one stand to maximize the chance of earning a free entry?

I will put up the solutions on this blog tomorrow. First, you need to make a whole-hearted attempt.

Nalin Pithwa.

Message for students: Ms. Sumita Mukherjee, Principal, Ryan International School, NOIDA

“You can lose everything, but not education” :

Ms. Sumita Mukherjee, Principal, Ryan International School, NOIDA, talked to DNA, (Mumbai, print edition, May 19 2017) about  the growing concern of peer pressure among young students, the importance of sex education in schools and more: Excerpts from the interview:

How can one help students struggling with peer pressure?

There are a lot of students dealing with performance and peer pressure. It’s very common among teenagers. First of all, we need to identify such students in our schools, and then understand their issues. Recently, we identified a Class 12 student in our school who was brilliant till Class 11. He suddenly stopped coming to school regularly. We called up his parents and what we got to know that the excuse he had given to them was that nothing important was happening at school. We asked the parents how they can take it for granted. We talked to him and realized that he was going through peer pressure. We told him about the challenges of life and convinced him that he can’t give up on his future like that. Now, he attends school regularly. So, we need to handle such students very carefully and it is also the responsibility of parents to approach the school immediately when something like this happens.

\vdots

Any message for students?

You can lose everything in life, but not education.

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Shared by Nalin Pithwa.

 

 

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