BR Maths Corner-1

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Vayutuvan
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Re: BR Maths Corner-1

Post by Vayutuvan »

matrimc wrote: 13a = 39 n + 2 x
where
3a - 2b = 13 n
2a + 3b = x
a an integer
SaiK: now you can see the one in red can be written as follows

13 (a-3n) = 2x

RHS is divisible by 13 so LHS has to be divisible by 13. But 13 and 2 are co-primes. So applying Euclid's lemma (p is 13, a is 2 and b is x) we get 13 | x which is 13 | (2a + 3b).
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Re: BR Maths Corner-1

Post by Vayutuvan »

AmberG: That is a nice a problem

Let me restate with subscripts and do some linear algebra with underlying finite fields.

3 x_1 - 2 x_2 = r_1
2 x_1 + 3 x_2 = r_2

There are two ways to look at this.

The vector r = [r_1 r_2] should lie in the column space of the matrix

Code: Select all


+---        ---+
|  3        -2  |       = A
|  2         3  |
+---         ---+

and the columns of A are orthogonal. So when we A' A is the diagonal diag(13,13).

So the solution of A' A x = A' b should lie in the field F_13^2 as A' A contains the unit of F_13^2.
another is to by simple geometry.

Take a grid and mark points at the intersections of the grids. Two vectors represented by points (3,2) and (-2,3) are drawn. Using those vectors as units color all the points say red. Now color all the points (13m,n) on the grid say with green. Then the intersection points which are colored yellow (colored both red and green) are the solutions. Since the first coordinate is in the F_13 the second also has to be in F_13.

There seems to be a connection to Quadratic residue. Am I correct?
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Re: BR Maths Corner-1

Post by Amber G. »

Big breaking news...

Counterexample to Fermat's Last Theorem Found!!!

There has been a really amazing development today on Fermat's Last Theorem. Noam Elkies has announced a counterexample, so that FLT is not true after all! .

The solution to Fermat that he constructs involves an incredibly large prime exponent (larger that 10^20), but it is constructive. The main idea seems to be a kind of Heegner point construction, combined with an really ingenious descent for passing from the modular curves to the Fermat curve. The really difficult part of the argument seems to be to show that the field of definition of the solution (which, a priori, is some ring class field of an imaginary quadratic field) actually descends to Q. I wasn't able to get all the details, which were quite intricate...
So it seems that the Shimura Taniyama conjecture is not true after all. The experts think that it can still be salvaged, by extending the concept of automorphic representation, and introducing a notion of "anomalous curves" that would still give rise to a "quasi-automorphic representation".
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Re: BR Maths Corner-1

Post by SaiK »

C/F from nukkad discussion: posting the article in full (just in case chindu dies earlier than BR).

Dated article!
For a career in mathematics
http://www.thehindu.com/sci-tech/scienc ... 747091.ece
S. Kesavan

Image
In India a mathematical career has been regarded as being synonymous with a teaching career, but a trained mathematician can be very well employed outside academia. Photo: S.R. Raghunathan

It is indeed possible to build a perfectly satisfying career in mathematics if one is deeply interested in the subject
India has a long and ancient mathematical tradition. The Sulvasutras, Vedic texts for the construction of ritual altars, contain a lot of geometrical results and constructions. These include a statement of the Pythagoras Theorem, an approximation to the value of ‘pi', and the ratio of the circumference of a circle to its diameter. India gave the world the decimal place value system, the modern way of writing numbers, and above all, the number ‘zero.' It boasts of mathematical schools like those of Aryabhata and Bhaskara. Much later, in the 15th century, came the flourishing School of Madhava in Kerala, which anticipated, by more than 200 years, several results of the Calculus invented by Newton and Liebniz.

There was a complete break in this tradition during the years of colonial rule. In the 20th century, perhaps inspired by Ramanujan's life, there was a revival, especially in the south, of mathematical research. In the post-Independence era, the Government of India established some schools of excellence, where several individuals distinguished themselves, and continue to distinguish themselves, by doing excellent work.

Nevertheless, for a country of India's size, despite having a large scientific workforce, we have failed to make the kind of international impact that countries like, say, China, have made. India's own scientific leaders have often bemoaned the ‘ocean of mediocrity' that has been created.

The main problem is that a mathematical career has been regarded as being synonymous with a teaching career. We religiously teach our children slokas like Guru Brahma, Guru Vishnu, Guru Devo Maheshwarah. However, equally cruelly and callously we say things like vakkillathavanukku vathiyar velai (a teacher's vocation is for those who have no other option). This has become a self-fulfilling prophecy of sorts. Barring a minuscule number of exceptions, India's brightest minds are not engaged in scientific research. The situation in general is that those who fail to join professional courses leading to gainful employment come to research as a last resort. These are the ones who will become the (uninspiring) teachers of the future — and we are caught in a vicious cycle.

The situation should, in reality, be the opposite. Those taking to a research career should be those who are passionately involved in the subject. As the experience of the information technology industry shows, anybody with a reasonable degree can be trained on the job and be well-employed, whereas that is not the case in academia.

It is indeed possible to build a perfectly satisfying career in mathematics (and much of this applies to other pure sciences as well) if one is deeply interested in the subject.

Job scene
Look at the job scene. A trained mathematician can be very well employed outside academia. Government departments engaged in space research (the Indian Space Research Organisation, or ISRO), defence research (Defence Research and Development Organisation, or DRDO), aeronautical research (National Aeronautics Limited, or NAL), all employ mathematicians to solve their special problems. Today, cryptology is in vogue (the systems ensuring the safety of your credit card transactions are based on some very sophisticated mathematics). Organisations such as the DRDO and the Society for Electronic Transactions and Security (SETS) are interested in mathematicians with training in this area. Financial mathematics is another area that leads to well-paid jobs. Computer giants such as IBM and Microsoft have research departments which have highly paid scientists who are either mathematicians or theoretical computer scientists. (They can, for all practical purposes, be considered as mathematicians). Thus, there is plenty of scope, outside academia, for well-paid jobs for mathematicians.

Having said this, it must be emphasised that the majority of mathematicians will end up in academic jobs, namely, in research and teaching.

What are the plus points of such a vocation?

•In India, all these jobs are in universities or in public-funded research institutions. With the implementation of the recommendations of the Sixth Pay Commission, the salary is nothing to be sniffed at. The entry point (roughly between the ages of 28 and 32) is that of an Assistant Professor, who can expect to start at a monthly basic salary of Rs.30,000. To this, add the dearness allowance (which has well crossed 502 per cent of the basic), transport allowance, and (if accommodation is not provided by the employer) a house rent allowance (which touches 30 per cent of the basic in the metros). Thus, before tax, we arrive at something like Rs.50,000 or more a month. This, unlike in industry, is not the ‘cost to company' but what the employee actually gets. Added to this are perquisites such as comprehensive health care, leave travel concession, aid to children's education and employer's contribution to the provident fund or the pension fund. All in all, the remuneration today does guarantee a very good standard of living with all the creature comforts.

In order to attract young Ph.D.s who have done rather well by way of research, especially but not limited to those from abroad who seek employment in India, the Department of Science and Technology (DST) offers the Ramanujan Fellowship for three years. It carries a high salary and a generous contingency grant that allows purchase of research equipment, travels abroad for conferences, and so on. Institutions like the IITs and the Indian Institute of Science (IISc) in Bangalore also offer generous start-up grants to freshly-recruited faculty members to facilitate their research.

•Job security.

•Job satisfaction: you get to choose your research problems.

•A good quality of life: the timings are regular with vacation periods that are well-defined.

•Plenty of opportunities to set up research collaborations with fellow-researchers in India and abroad, providing possibilities of interesting domestic and international travel.

•Being in contact with young minds all the time has a rejuvenating effect on one's outlook to life.

On the other hand, one should ensure that one is really interested in the subject. To rise in the profession one needs to have a reasonably steady research output for nearly three to four decades. The real downside is that the gestation and apprenticeship period is quite long. It takes about five years to get a master's degree and between three to five years more for the doctoral degree. Even after that, it is expected that a person does at least two years of post-doctoral work, which is the time when one emerges from the shadows of the thesis supervisor and chalks out one's own path of research. Thus, as mentioned earlier, one can expect to get one's first job when in the 28-32 age group. But this period is not financially barren, and the remuneration keeps increasing.

Job opportunities
What about job opportunities in India? There are three kinds of institutions of higher learning.

Purely research-oriented institutions like the Tata Institute of Fundamental Research (TIFR) in Mumbai, the Institute of Mathematical Sciences (IMSc) in Chennai, and the Harish Chandra Research Institute (HRI) in Allahabad. Interestingly, all these are autonomous aided institutions that are fully supported by the Department of Atomic Energy (DAE) of the Government of India. TIFR is now a deemed university, while the IMSc and the HRI are affiliated to the deemed university called the Homi Bhabha National Institute (HBNI) that covers all other aided institutions of the DAE.

Institutions of teaching and research which can offer degrees but do not come under the purview of the University Grants Commission (UGC). These are set up by Acts of Parliament, and some come under the Ministry of Human Resource Development (MHRD). These are the Indian Statistical Institute (ISI), the IITs, the IISc, and the newly set up Indian Institutes of Science Education and Research (IISERs) in Bhopal, Kolkata, Mohali, Pune and Thiruvananthapuram, and the National Institute of Science Education and Research (NISER) in Bhubaneswar (set up by the DAE). Then there is the precursor to these latter new institutes, the unique Chennai Mathematical Institute (CMI), which is an example of public-private partnership. ISRO has also established its own such institution in Thiruvananthapuram.

The State and Central universities. While the State universities have plenty of vacancies, these being filled is often tied to the policies and politics of the State governments.

All the other institutions of research and teaching mentioned above have well-established and transparent methods of selection, and all of them have a crying need for fresh faculty. In fact, the need is so great and the supply so meagre that the age of retirement has been increased to 65 for these institutions. And in many cases they are allowed to re-employ superannuated faculty members till they are 70. The government has suddenly started NISER, the five IISERs and about eight new IITs, all of which need faculty members. These are currently functioning with a bare minimum of recruits, augmented by adjunct faculty members, who are retired mathematicians. This is not sustainable in the long run.

Thus, for those who hold a reasonably good doctoral degree, there are plenty of job opportunities in such institutions. This will be so for a long time to come.

Even the existing institutions like the IITs face continuous attrition due to retirement of faculty members who were engaged from the 1960s onwards.

Training
Now for the training process of a mathematician in India. The regular route for a student is a three-year B.Sc. course followed by a two-year M.Sc. programme in mathematics, after which she or he could join a doctoral programme in a recognised university or research institution. There are the following variants to this theme.

The IIT-Kanpur pioneered the five-year M.Sc. programme (admission is through the joint entrance examination) which combined the B.Sc. and M.Sc. programmes. IIT-Bombay followed suit. Now, this pattern is followed by all the IISERs and NISER. The Central University of Hyderabad and that of Pondicherry have also started such programmes.

Recently, the three science academies in India have been advocating educational reform that involves the introduction of a four-year B.S. programme followed by a year of research and training leading to an M.S. The IISc will launch the first such programme in August 2012.

Institutions of pure research (the TIFR, the IMSc and the HRI), the IISc and the CMI also have integrated Ph.D. programmes. Promising students are selected after a bachelor's degree in any science discipline or engineering directly for their Ph.D. programmes, provided they clear the (very rigorous) entrance tests and interviews on a par with M.Sc. candidates. They pick up an M.Sc. degree after two initial years of course work and research.

All the IITs and universities also have independent M.Sc. and Ph.D. programmes. Admission is based on entrance tests and/or interview. The CMI has an M.Sc. programme in applications of mathematics with specialisation in financial mathematics and computational applications of mathematics. It is contemplating a stream specialising in cryptology. The ISI has an M. Math. Programme, held alternatively at its Kolkata and Bangalore campuses.

A special word on the undergraduate programmes of the CMI, which is B.Sc.(Hons.) in Mathematics and Computer Science in Chennai, and the ISI — B.Math at its Bangalore campus. These are not for the faint-hearted. But if a student has a strong taste and talent for mathematics from an early age, these are the places to go for mathematics education. Both these programmes are very intense. At the end of three years, the students can compete with any master's level student anywhere on equal terms — and often they fare better.

As a measure of the success of these programmes, it must be said that their graduates have managed to breach the U.S. firewall that requires a four-year collegiate-level education to enter graduate school, by being directly admitted, with full aid, to graduate schools such as Caltech, Chicago, Princeton, MIT (and so on in the U.S.), the Max Planck institutes in Germany, and the elite Ecole Normale Sup´erieure in France, after finishing the three-year degree programme. Students from the early batches have started completing their doctorate work and are already making a mark. It is gratifying that some have come back to take up positions in India.

Finally, about scholarships. First of all, there is the Kishore Vaigyanik Pratsohan Yojana which conducts a test for high school students. The successful ones opting for a career in science get a handsome scholarship all through their higher education, up to completion of the doctoral programme. The CMI and the ISI provide modest stipends to their undergraduates and postgraduates, together with tuition waiver, as long as the students maintain a healthy academic performance.

For the doctoral programmes, university students need to take an examination conducted by bodies like the Council of Scientific and Industrial Research or the DST for a research fellowship. The current rates are Rs.16,000 for the first two years and, subject to satisfactory performance, Rs.18,000 a month thereafter. There is an annual contingency grant as well.

All research institutions and institutions of teaching and research mentioned here have their own funding for Ph.D. scholarships at the same rates. In case the institution cannot provide subsidised accommodation on campus, house rent allowance at the same rates as applicable to faculty members is allowed.

Post-doctoral fellowships provide for a consolidated pay ranging from Rs.21,000 to Rs. 25,000 a month (with the provision for HRA), along with a contingency grant, depending on the candidate's post-doctoral experience.

The National Board for Higher Mathematics (NBHM), set up by the DAE to promote mathematics, conducts an examination every year for the award of a scholarship for M.Sc. programmes in mathematics in any recognised university or institution, and pays a monthly stipend of Rs.6,000. The advertisement appears in newspapers by the end of June; the written test is usually held towards the end of September. It also awards Ph.D. scholarships, at the same rates as other research fellowships, by conducting another examination which is advertised in November; the test is usually by the end of January or early February. The NBHM also offers post-doctoral fellowships.

To sum up, if a student has the taste and the talent for mathematics, it is possible to make a satisfying, interesting, respectable and remunerative career out of it. If you think you have it in you, just go for it. Study abroad if you really want to; it can broaden your horizons. But do come back to inspire future generations so that India will become a mathematical superpower in the coming decades.

Parents ought to let children do whatever they are best suited for — literature, dramatics, mathematics, painting and so on. They should not try to live out their ambitions through them. While it may be a status symbol to count a non-resident Indian in the family, as one grows older there is a pleasure and sense of security in having one's children living and working close by.

(The author is a Professor of Mathematics at the Institute of Mathematical Sciences, Chennai.)

Keywords: Srinivasa Ramanujan, Ramanujan birth anniversary, mathematics as career
ArmenT
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Re: BR Maths Corner-1

Post by ArmenT »

I saw a very elegant proof for sum of natural numbers = -1/12 that I hadn't seen before, so I thought I'd share with you all.

Let X = 1 + 2 + 3 + 4 + 5 + ....
Therefore 4*X = 4 + 8 + 12 + 16 + ....

Then, we compute X - 4X. We can arrange the terms like this:

Code: Select all

X - 4X = 1 + 2    + 3 + 4   + 5 + 6 + .....
             -4        -8        -12   ... 
or, rearranging a bit:

Code: Select all

X - 4X = 1 - 2 + 3 - 4 + 5 - 6 +....
Now, the right hand side of this infinite series evaluates to 1/4, as proved by Leonhard Euler many years ago.
Therefore, X - 4X = 1/4
or -3X = 1/4

Hence, X = -1/12

Therefore, sum of natural numbers is -1/12
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Re: BR Maths Corner-1

Post by Amber G. »

^^^ :)
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Re: BR Maths Corner-1

Post by Bade »

How evil of math guys to not post this too. :-) http://scitation.aip.org/content/aip/ma ... /PT.5.8029

This number 12 comes up in many contexts, for fixed width detector strips of say 'p' in width, the spatial resolution is p/sqrt(12) !
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Re: BR Maths Corner-1

Post by Amber G. »

ArmenT and Bade...

Let me also add, John Baez's rather recent proof ..
http://math.ucr.edu/home/baez/qg-winter2004/zeta.pdf

BTW, John Baez (A physicist , a famous quantum gravity) is cousin of Joan Baez (my favorite singer)... famous folk singer and activist. (Joan Baez's father as well as uncle both are pretty well known physicist).
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Re: BR Maths Corner-1

Post by Vayutuvan »

ArmenT is late by about 19 days.

http://en.wikipedia.org/wiki/Bolzano%E2 ... orem#Proof

By the way there is more fun to be had with this one. :)

Just in case people didn't get it - since yesterday was Easter one has to go Easter egg hunting from that link to convergent/divergent series
Last edited by Vayutuvan on 22 Apr 2014 10:12, edited 1 time in total.
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Re: BR Maths Corner-1

Post by SaiK »

math gurus, please get me some head start on DSM? moorkh examples onlee. tia
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Re: BR Maths Corner-1

Post by Vayutuvan »

Saik DSM :?:
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Re: BR Maths Corner-1

Post by Amber G. »

SaiK
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Re: BR Maths Corner-1

Post by SaiK »

:rotfl:

design structure matrix.
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Re: BR Maths Corner-1

Post by Amber G. »

I liked it:

RSA:

There once was a girl from Great Bend
Who wanted to hear from a friend.
But to her disgust
There was no way to trust
The network, beginning to end.
She needed a way to encrypt
Message X before it was shipped.
"It needs to be true
That it's easy to do,
But hard to do backwards," she quipped.
"I can multiply numbers with ease—
Numbers as large as you please.
But to go in reverse
Is very much worse.
So I'll pick two big primes to make keys.
"One is p, other's q. Next step, then,
Is to find out their product, called N.
Now choose e: This time,
It's a relative prime
To the Euler φ-function of N.
"This function is easy to do:
Find pq less p, minus q,
Plus one. So we see
It's not hard to find e.
Now the last thing we need is called μ.
"In Greece, long ago, Euclid taught
A way to find μ, which is sought,
So that 1 less μe
Will turn out to be
A multiple of φ(N) less than nought.
"So I have μ and e. Now I may
Keep μ private, but give e away.
My friend gets the key,
Computes X to the e
Mod N, and sends that my way.
"This new message Y seems complex.
To decrypt it, what to do next?
Well, all that I do
Is find Y to the μ
Mod N; the result will be X.
"This encryption is hard to undo
Without knowing q, p, or μ.
The 'security guard'
Is that factoring's hard,
Unless there's a brilliant breakthrough."
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Re: BR Maths Corner-1

Post by ArmenT »

A Beautiful Mind: Brain Injury Turns Man Into Math Genius
In 2002, two men savagely attacked Jason Padgett outside a karaoke bar, leaving him with a severe concussion and post-traumatic stress disorder. But the incident also turned Padgett into a mathematical genius who sees the world through the lens of geometry.

Padgett, a furniture salesman from Tacoma, Washington, who had very little interest in academics, developed the ability to visualize complex mathematical objects and physics concepts intuitively. The injury, while devastating, seems to have unlocked part of his brain that makes everything in his world appear to have a mathematical structure.
The article is a very interesting read...
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Re: BR Maths Corner-1

Post by Amber G. »

ArmenT wrote:A Beautiful Mind: Brain Injury Turns Man Into Math Genius
....
The article is a very interesting read...
FWIW: I read that article, and other pieces as there has been lot of media attention on him. I remain unimpressed as I have not seen a single point which shows any evidence of "genious" in that person. To me, it seems like people are very gullible when it comes to math, and even some mumbo-jumbo can make a news story. (eg "He started sketching circles made of overlapping triangles, which helped him understand the concept of pi, the ratio of a circle's circumference to its diameter. There's no such thing as a perfect circle, he said, which he knows because he can always see the edges of a polygon that approximates the circle...(so? - what math concept is here? - or talking about "Planck length" - which involves nothing more than repeating something read from wiki...)
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Re: BR Maths Corner-1

Post by Multatuli »

GeoGebra

Dynamic mathematics & science for learning and teaching

* From elementary school to university level
* Interactive geometry, algebra, statistics, and calculus software
* Tens of thousands of free materials

http://www.geogebra.org/cms/en/
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Post by Amber G. »

Wonder if anyone watched Mathcounts countdown round competition on ESPN3 this Friday..

(This year the Raytheon MATHCOUNTS National Competition took place in Walt Disney World..IMO more fun than Washingron DC, where some of the past events I attended :)

Swapnil Garg of Sunnyvale, Calif was the top winner. California team was first too. (Our state team did okay too)...

Okay, here is the final question from the countdown round .. (Countdown round is at the end of competition where top 12 contested - from written exam - try out how fast they can solve the problem. This part is fun to watch)

Start a stop watch before reading further...
The smallest integer of a set of consecutive integers is -32. If the sum of these integers is 67, how many integers are in the set?
Stop the watch when you answer the question..:)



Swapnil answered this correctly in 12 seconds - after the Q was flashed on his screen beating the other buzzer... Another interesting part was that he never touched his scratch paper with his pencil answering any of his questions..
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Re: BR Maths Corner-1

Post by Vayutuvan »

Wow that is fast. By the time I understood and formulated an approach, it was more than 7 or 8 seconds. It is 67 of course - not to forget 0.
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Re: BR Maths Corner-1

Post by ArmenT »

Got it in about 4 secs. Trick was to just confirm that the last two integers added up to 67 in my head.
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Re: BR Maths Corner-1

Post by Vayutuvan »

ArmenT, that is fast. We have a few guests for lunch and I posed this question. My friend took about 7-8 seconds and came back with " do I include nothing (Hindu shunya). As I suspected for long I am a little slow on the take which explains why I am quite bad at standardized tests. I lies track of my thought process mid sentence some times. :)

Here is a generalization

If there are (2n+3) consecutive numbers which add upto (2n+3) what is the starting number? :)
Last edited by Vayutuvan on 12 May 2014 10:14, edited 3 times in total.
Multatuli
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Re: BR Maths Corner-1

Post by Multatuli »

My Guess would be -32. Don't ask for a systematic approach to the solution, I am just "reverse engineering" the final Mathcounts question.

A "systematic approach" would be try out -2, -3 etc.

Added later:

Just write a nice little proggie to do it.
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Post by Amber G. »

Okay how about this.. From the same competition... (6th-8th grade students)
A circle passes through two diagonally opposite vertices of a 3-inch by 4-inch rectangle. What is the least possible distance between the center of the circle and a vertex of the rectangle?
The following was interesting too, as this was answered by fewest number (less than 6% answered it correctly) of students..

(Mind you these are top (among the brightest) about 200 students nation wide - selected from about a few hundred thousand students) - you have about a minute to solve)
Larry tells Mary and Jerry that he is thinking of two consecutive integers from 1 to 10. He tells Mary one of the numbers, and he tells Jerry the other number. Then the following conversation occurs between Mary and Jerry:
Mary: I don't know your number.
Jerry: I don't know your number, either.
Mary: Ah, now I know your number.
Assuming both Mary and Jerry used correct logic, what is the sum of the possible numbers Mary could have?
Last edited by Amber G. on 12 May 2014 10:07, edited 1 time in total.
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Re: BR Maths Corner-1

Post by Vayutuvan »

Multatuli wrote:My Guess would be -32. Don't ask for a systematic approach to the solution, I am just "reverse engineering" the final Mathcounts question.

A "systematic approach" would be try out -2, -3 etc.

Added later:

Just write a nice little proggie to do it.
Actually it is -n.

-n to +n including 0 add to 0 but those are only 2n+1 numbers. Now add 2 numbers after n which are n+1 and n+2 which add up to 2n+3. Also the set cardinal it is 2n + 3 as well.
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Re: BR Maths Corner-1

Post by Vayutuvan »

Amberg 2.5 and 11?

My son was in the school team. They did reasonably well but could not make it to the state.
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Re: BR Maths Corner-1

Post by ArmenT »

Amber G. wrote: The following was interesting too, as this was answered by fewest number (less than 6% answered it correctly) of students..

(Mind you these are top (among the brightest) about 200 students nation wide - selected from about a few hundred thousand students) - you have about a minute to solve)
Larry tells Mary and Jerry that he is thinking of two consecutive integers from 1 to 10. He tells Mary one of the numbers, and he tells Jerry the other number. Then the following conversation occurs between Mary and Jerry:
Mary: I don't know your number.
Jerry: I don't know your number, either.
Mary: Ah, now I know your number.
Assuming both Mary and Jerry used correct logic, what is the sum of the possible numbers Mary could have?
5 or 17?
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Re: BR Maths Corner-1

Post by Patni »

Amber G. wrote:Okay how about this.. From the same competition... (6th-8th grade students)
A circle passes through two diagonally opposite vertices of a 3-inch by 4-inch rectangle. What is the least possible distance between the center of the circle and a vertex of the rectangle?
The following was interesting too, as this was answered by fewest number (less than 6% answered it correctly) of students..

(Mind you these are top (among the brightest) about 200 students nation wide - selected from about a few hundred thousand students) - you have about a minute to solve)
Larry tells Mary and Jerry that he is thinking of two consecutive integers from 1 to 10. He tells Mary one of the numbers, and he tells Jerry the other number. Then the following conversation occurs between Mary and Jerry:
Mary: I don't know your number.
Jerry: I don't know your number, either.
Mary: Ah, now I know your number.
Assuming both Mary and Jerry used correct logic, what is the sum of the possible numbers Mary could have?
25
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Re: BR Maths Corner-1

Post by Amber G. »

matrimc wrote:Amberg 2.5 and 11?

My son was in the school team. They did reasonably well but could not make it to the state.
I think both answers are incorrect..
First one is easy enough. (Think of the rectangle coordinates are: (-2.5,0),(0.7,2.4),(2.5,0)(-0.7,2.4) (Easy to see that in (3,4,5) (ABC when AC=5) right triangle, if you draw a perpendicular from B to AC it is 12/5=2.4) .. The center has to be on the y axis so answer is 0.7 :)

For the second, 11 was one of the most popular answer (more than half got that answer) but it is not correct..
(And no, it is not the other answers posted here either :) (at least the official answer )
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Re: BR Maths Corner-1

Post by Amber G. »

Hi Matrimc - I hope your son likes these problems... (It seems there is some interest here too..)
So One last but one typical problem from the sprint round (you have about a minute to solve)
Two random integers, a and b, are independently chosen, with replacement,
from 1 to 1000, inclusive. What is the probability that both 2^a + 2^b and 3^a + 3^b are multiples of 5?
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Re: BR Maths Corner-1

Post by Amber G. »

Okay I just got an interesting email from one of the young 6th grader bright girl .. ( from the team I coached this year)..
I forgot to mention. We had the exact same triangle challenge problem you gave to us in our team round. One of us even remember the correct solution and we got the answer right.
(I got a very nice letter BTW, and glad that many young kids are interested in math)

BTW, IIRC the exact triangle problem appeared here in BRF (by yours truly) a few years ago..this kind of problem is one of my favorite geometry problem (which even using trig is not easy but still not too hard, if you hit the right idea)

Here is the problem again.. (This was the last problem of the team round)
Image

(Team round, you have more time, and the whole team (4 people) can work on it .. as difficulty go, this was supposed to be the hardest problem)
Last edited by Amber G. on 13 May 2014 01:35, edited 4 times in total.
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Re: BR Maths Corner-1

Post by Vayutuvan »

AmberG, these kids are good. Thanks for sharing the problems. I will pose them to my son.
The circle problem I made the obvious mistake (assuming the circle to have the diagonal as the dia of course it is the digonal). My mistake is not drawing a diagram :((
Last edited by Vayutuvan on 13 May 2014 01:43, edited 1 time in total.
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Re: BR Maths Corner-1

Post by Vayutuvan »

Multatuli and others, by the way substitute a big number for n like 1,242,764 or even a billion or some such and give the problem. People will take time to reverse engineer the formula. Abstraction and closed form solution (of course works only when it is a "simple function" to compute, i.e. the computational complexity of the function is low). If the numbers become "very very large" then the program needs to use linked lists because even 128 bit (long long int) would not be able to store large numbers (unless one writes the program in Python or Scheme where the large numbers are built in).

There are two notations for large numbers - one by Knuth and another by John Conway - both of which can be found in WIkipedia. Knuth uses an up arrow notation and Conway uses arrow notation. Conway's notation can express numbers many times larger than Knuth's notation. Some interesting things about the conversions from one notation to the other are listed at the Wikipedia. Conway is one mathematician who is simply brilliant.
Last edited by Vayutuvan on 13 May 2014 06:36, edited 1 time in total.
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Re: BR Maths Corner-1

Post by Multatuli »

AmberG, I did it with analytical geometry, and the minimal distance is the distance between the points:

D=(0,4) and F=(-21/50,86/25), that is 0,69 or 0,7.

(Rectangle with vertices A=(0,0), B=(3,0), C=(3,4), D=(0,4), the circle passes through A and C, so the center of all the circles must lie on y=-3*x/4 + 25/8.)

My answer for the second problem (uber murkh answer, I just add all integers from 1 to 10): 55.

I know that the second is probably incorrect, so please explain the reasoning in detail, otherwise I just won't get it.

And yes, these kids are very, very good, I couldn't do the first problem with synthetic geometry, but these kids did it in under one minute!
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Re: BR Maths Corner-1

Post by Vayutuvan »

Multiatuli

You have to assume that M does not have the end numbers 1 or 10. Otherwise she would have known J's number. So M has 2 between 2 and 9. Now J says he does not know M's number. So J does not have 1 and 10 either. This is where I goofed and gave 11, which is wrong. It could be 44. But we have one more data point that M says that she knows J's number now. Can 44 be reduced further? Still thinking...
Last edited by Vayutuvan on 13 May 2014 03:09, edited 1 time in total.
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Re: BR Maths Corner-1

Post by Amber G. »

Multatuli wrote:AmberG, I did it with analytical geometry, and the minimal distance is the distance between the points:

D=(0,4) and F=(-21/50,86/25), that is 0,69 or 0,7.

(Rectangle with vertices A=(0,0), B=(3,0), C=(3,4), D=(0,4), the circle passes through A and C, so the center of all the circles must lie on y=-3*x/4 + 25/8.)
Good, this way is not too bad (and fast enough) but the answer is exactly 0.7
Note that (4-86/25 = 14/25=28/50) and we have 21^2+28^2 = 7^2(3^2+4^2) = 7^2*5^2 etc..
(In other words, a point equidistant from (0,0) and (3,4) is given by
x^2+y^2 = (x-3)^2+(y-4)^2 which simplifies to your equation 6x+8y-25=0
and the required distance from (0,4) is (6*0+8*4 -25)/sqrt(6^2+8^2) which is 7/10.
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Re: BR Maths Corner-1

Post by Vayutuvan »

AmberG it is 22 and Mary could have one of the numbers {2,3,8,9}.
So we have to go through each round and eliminate numbers for M and J.

First Round: M doesn't know J s number -> M = {2-9}, J = {1-10}
Second Round: J doesn't know M's number -> M = {2-9}, J = {3-8}
Third round: M knows J's number -> The function from M's number to J's number M finds is {(2,3),(3,4),(8,7),(9,8)}

If in the third round M did not know J's number then M's number could one of {4,5,6,7}, i.e. again the sum is 22.

Third round whether Mary knows or not is important to reduce the set sum to 22. If Mary doesn't say out loud whether she knows J's number or not, then we cannot reduce the sum from 44 to 22.
Last edited by Vayutuvan on 13 May 2014 06:31, edited 1 time in total.
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Re: BR Maths Corner-1

Post by Amber G. »

^^^ Yes the answer was 22.. (as you have explained)
(1 and 10 is not possible (for anyone) (else one will know the others number),
(4,5,6,7) is not possible for M.
-- because then one can not deduce other's number (" Ah, now I know your number" will not be possible).. so you are left with (2,3,8,9) and this sum is 22.
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Re: BR Maths Corner-1

Post by Amber G. »

I know that the second is probably incorrect, so please explain the reasoning in detail, otherwise I just won't get it
.

FWIW, expanding on what has already been said...

Remember numbers are two consecutive integers between 1 and 10, and M&J are both employing perfect logic.

[quote]Mary: I don't know your number. (statement 1)
Jerry: I don't know your number, either. (statement 2)
Mary: Ah, now I know your number. (statement 3) [/quote]

After statement 1, Jerry (and everyone else) know(s) that Mary's number is NOT 1 (or 10) (because M will then know J's number as 2 is the only number consecutive to 1)

So only possible number Mary can have are (2,3,4,5,6,7,8,9)

After statement 2, Mary (and everyone else) know(s) that Jerry's number is neither 1, nor 2 (or 9 or 10 on the other end).. (because now 2 (and 9) have only one neighbor)

So possible numbers for Jerry are (3,4,5,6,7,8)

Now after statement 3, only possibilities are Mary has 2 (and Jerry 3 ) , or Mary has 3 (and Jerry has 4) and similar pairs on the other end (8 or 9 )..
(For example if Mary had 4 she can not tell J's number because it could be either 3 or 5)
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Re: BR Maths Corner-1

Post by ArmenT »

^^^
Aah, I was wondering about that, because I didn't see a flaw in my logic at all (was going to post my logic after getting back from work, but it is the same as yours), but I posted the two possibilities (5 and 17) rather than the sum of them both (5 + 17 = 22). I interpreted the question to be the sum of each of the two possibilities, rather than the total sum :oops:
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