Mathematics Hothouse

Maths in a minute! Thanks to PlusMath!

June 23, 2017 – 2:02 pm

https://plus.maths.org/content/maths-in-a-minute?nl=0

By nkpithwa | Posted in IITJEE Foundation Math IITJEE Main and Advanced Math and RMO/INMO of (TIFR and Homibhabha) | Comments (0)

All 30 candidates of Patna’s Super 30 crack IITJEE Advanced 2017!

June 17, 2017 – 2:40 pm

(This piece of news is also a bit dated, but v v inspirational)

(Reference: DNA newspaper, print edition, Mumbai, June 17 2017, Education section)

Patna: Mathematician Anand Kumar’s Super 30 has once again created a record as its 30 out of 30 candidates cracked the IITJEE Advanced Examination (2017) results which were announced last Sunday.

“I am happy that my 30 out of 30 candidates have cracked IITJEE Advanced this year. Now time has come to give Super 30 some much needed expansion. We will now organize tests for selecting students in different parts of the country and give all the details on the website,” the Super 30 founder said.

Talking to media persons after the results, Kumar attributed his yet another success story of Super 30 to the hard work and perseverance of his students and said that its time to expand his famous Super 30 in the country.

Kumar’s Super 30 provides free coaching along with lodging to students of underprivileged sections of the society.

Super 30 has completed its 15 years of journey during which it has sent 396 out of 450 candidates to IIT. Kumar said adding that it coached 450 candidates in 15 years, 30 candidates every year.

Among the 30 students of Super 30 who have cracked prestigious IITJEE Advance 2017, there are wards of landless farmer, egg seller, unemployed father.

There are several inspiring tales that have emerged from the Super 30 coaching institute this year too.

Be it the story of Kevlin, son of an unemployed father, or roadside egg seller’s son Arbaaz Allam, a farm labourer’s son Abhishek, they all stand tall having overcome the stiff odds of poverty and deprivation to make it to IIT and become an inspiration for several others like them.

Kevlin’s father Deepak is unemployed. He teaches yoga to people, but still his earning is not enough to make both ends meet. But Deepak knew that the only way to get out of the life of poverty is through education. Tears of joy continue to roll down his cheeks as he sits with mathematician Anand Kumar.

“I had heard of Super 30 almost 10 years ago and since that day I dreamt of seeing my son here to realize my dream. Today my son has realized it,” said Deepak.

Arbaaz’s father sells eggs in Bihar’s Biharsharief district. But he never lost heart. He always aspired to be at the top to change the course of his life.

“Anand Sir made me feel confident about my abilities. He boosted my confidence. Now, I think it is a matter of a few years when my father will not have to sell eggs braving the chilly winds of winter nights. I will also have a house where I will live with my father and mother,” said Arbaaz.

Kaushlendra Kumar of Nalanda is a landless farm labourer. Kaushlendra is proud of his son Arjun Kumar and Super 30. Arjun proudly recounts how his journey from a village school devoid of even basics took him to Super 30 and now he plans to step into IIT. “It is due to programmes like Super 30 that make even students from underprivileged sections dream of IIT. Now after engineering, I plan to go in for UPSC,” Arjun said.

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Hats off to Prof. Anand Kumar and his team !!! From Nalin Pithwa.

Petronet Kashmir Super 30: Nine crack IITJEE 2017 exams

June 17, 2017 – 1:45 pm

(This news is a bit delayed here on this blog, nevertheless, inspirational.)

Reference: DNA, Mumbai print edition, June 15 2017:

Srinagar: Petronet LNG Ltd. (PLL), which started its CSR voyage from western India and traversing South and East, have now reached Northern state of Jammu and Kashmir.

Considering the lack of facilities/faculty in J and K to impart coaching for Engineering Entrance Examination to facilitate admissions to IITs/NITs/other institutions of repute, Petronet LNG Ltd. sponsored 40 underprivileged students in 2016-17 to fulfill their dream of higher engineering education (35 boys and 5 girls) under Petronet Kashmir Super 30 programme in association with Indian Army, CSRL. This 11 months’ residential programme was attended by 40 students (kargil 7, Pulwama 5, Bandipura 6, Baramulla 4, Anantnag 7, Ganderbal 4, Kulgam 1, Tangmarg 2, Ramban 1, Shopian 1, Sopora 1, Nagrota 1).

Despite hardships in Kashmir and educational institutions being closed last year, the project continued with its mission. The declaration of IITJEE results on April 27 2017 saw selection of 28 students (including 2 girls) who were coached under Petronet Kashmir Super 30 cracking the IITJEE mains exams. Six other students will join Regional Engineering Colleges.

Shri Dharmendra Pradhan, Hon’ble Minister of State (I/C) Petroleum and Natural Gas and Mr. Chowdhary Zulfkar Ali ji, Hon’ble Minister for Dept. of Food, Civil Supplies and Consumer Affairs and Information Dept., J and K, interacted with students of Petronet Kashmir Super 30 at Delhi and felicitated them for their hard work and achievement.

Shri Pradhan expressed happiness that students from underprivileged backgrounds from places like Kupwara, Pulwama, Anantnag, Shopian etc. have shown their determination and got selected in prestigious institutes. He also announced on behalf of Petronet LNG Ltd. that 100 students will be sponsored from Kashmir for engineering entrance next year.

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🙂 🙂 🙂 Congratulations to Petronet (Kashmir) LNG Team including their faculties and the students from Nalin Pithwa !

By nkpithwa | Posted in IITJEE Foundation Math IITJEE Main and Advanced Math and RMO/INMO of (TIFR and Homibhabha) | Comments (0)

Who wants to be a mathematician

June 2, 2017 – 10:31 am

I wish that there were official programmes like the following in India too:

http://www.ams.org/programs/students/wwtbam/wwtbam

By nkpithwa | Posted in careers in mathematics, mathematicians, motivational stuff | Comments (0)

Something Cute I Never Noticed Before About Infinite Sums

May 31, 2017 – 7:21 pm

Source: Something Cute I Never Noticed Before About Infinite Sums

By nkpithwa | Posted in IITJEE Foundation Math IITJEE Main and Advanced Math and RMO/INMO of (TIFR and Homibhabha) | Comments (0)

Arithmetic Puzzle

May 28, 2017 – 6:34 pm

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…

View original post 35 more words

By nkpithwa | Posted in IITJEE Foundation Math IITJEE Main and Advanced Math and RMO/INMO of (TIFR and Homibhabha) | Comments (0)

APJ Abdul Kalam : How to manage success and failure

May 24, 2017 – 9:08 am

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…

By nkpithwa | Posted in motivational stuff | Comments (0)

Three in a row !!!

May 24, 2017 – 2:34 am

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.

By nkpithwa | Posted in Fun with Mathematics, Pre-RMO | Comments (0)

Follow Descartes’ Historically Famous Problems !

May 24, 2017 – 1:59 am

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.

By nkpithwa | Posted in co-ordinate geometry, IITJEE Mains | Tagged coordinate geometry | Comments (0)

Solutions to Birthday Problems: IITJEE Advanced Mathematics

May 23, 2017 – 9:04 pm

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.