The Vedic Maths Forum India Blog

Vedic Maths Journeys and Vedic Maths Online Pinboard

noreply@blogger.com (The Vedic Maths Forum India) — Wed, 25 Apr 2012 10:09:00 PDT

Vedic Maths has given me a new hobby! That of travelling....There are a lot of invitations which we get from various corners of the globe...and we love visiting these new places , sometimes at remote corners of the globe.

Its fun, exciting and lovely meeting new people in new cultures with diverse background all connected by a single thread of Vedic Mathematics!

So we invite travelogues and stories from various globetrotters who teach Vedic Maths....tell us about your travels and your experiences! Together we can then all learn and share each other's experiences.

This summer we are happy to launch our Online Vedic Maths PinBoard on Pinterest. Pinterest is a pinboard-style social photo sharing website that allows users to create and manage theme-based image collections such as events, interests, hobbies and more.

So we decided to share our snaps and some nice happy fun moments out here. Just click here to visit our Vedic Maths Online Pinboard.

So if you have some Vedic Maths related snaps which you want the world to see.....just send them in at gtekriwal@vedicmathsindia.org and we will publish it on our Pinboard with due credits to you. Happy Pinning everyone!

Its Summer Time!

noreply@blogger.com (The Vedic Maths Forum India) — Wed, 25 Apr 2012 09:51:00 PDT

Yippee! Its that time of the year again when cities and town become abuzz with Vedic Mathematics Workshops. Children are free from classes and teachers can organize Vedic Maths Summer Camps in various schools and institutions.

The Vedic Maths Community is very active in this phase now. So we thought we would extend out our helping hand to those people who are organizing Vedic Mathematics Workshops in various countries and cities. Entrepreneurs who need Vedic Maths Teachers support can just write in a mail to gtekriwal@vedicmathsindia.org and we will be happy to send our teachers to your city or town to conduct workshops.

This summer our teachers are travelling to the Middle East , Uttar Pradesh, Madhya Pradesh, West Bengal and even as far as Kerela. So entrepreneurs out there jump on this opportunity to organize Vedic Maths Workshops. The Vedic Maths Forum India will be happy to help with Marketing Materials and a solid plan to make your workshops successful.

Just write in to gtekriwal@vedicmathsindia.org and we will be happy to help you.

Mental Calculation World Cup 2012 Announcement

noreply@blogger.com (The Vedic Maths Forum India) — Wed, 04 Apr 2012 13:49:00 PDT

Mental calculators from all over the world are invited to the 5th Mental Calculation World Cup 2012.

Venue: Mathematikum, a science museum in Gießen, Germany

Date: 29 September - 1 October 2012

Registration:
The registration form can be found on the MCWC website (http://www.recordholders.org/en/events/worldcup/2012/).

Deadline for Registrations: 1 June 2012

More information can be found on the web site:
http://www.recordholders.org/en/events/worldcup/2012/

Squares of Sums and Sums of Cubes

noreply@blogger.com (Alex Greene) — Thu, 08 Mar 2012 10:32:00 PST

I recently stumbled upon a brief blog post which demonstrated an interesting phenomenon.

The square of the sum of the series of consecutive numbers from 1 is equal to the sum of the cubes of the numbers.

The article gave specific examples:-

(1+2+3)² = 1³+2³+3³

and

(1+2+3+4)² = 1³+2³+3³+4³

The above can, of course, be verified: (1+2+3)² = 6² = 1+8+27; (1+2+3+4)² = (1+2+3)² + 4³ = 36+64 = 100 = (1+2+3+4)².

This extends to (1+2+3+4+5)²; 1³+2³+3³+4³+5³ = (1+2+3+4)² + 5³ = 100 + 125 = 225 = (1+2+3+4+5)².

A satisfying blog post, and thought provoking. However, it stopped short of providing a general proof for an arbitrary natural number n.

Well, here goes.

I start with the assertion that there exists a function, f(x), which is expressed as:-

f(x) = 1+2+3+4+5+6+7 ...

Better expressed as:-

There must exist a function, g(x), such that

g(x) = (1+2+3+...)²

or

Finally, there exists a function, h(x), such that

h(x) = 1³+2³+3³+...

or

The aim of this article is to assert, for any arbitrary natural integer n, that

(1+2+3+...)² = 1³+2³+3³+...

or

For a given arbitrary integer n, if g(x) = h(x) for x=n, then

(1+2+3+...+(n-2)+(n-1)+n)² = 1³+2³+3³+...+(n-2)³+(n-1)³+n³

Since 1³+2³+3³+...+(n-2)³+(n-1)³ must be equivalent to (1+2+3+...+(n-2)+(n-1))²,

f(n) = (1+2+3+...+(n-2)+(n-1))² + n³

must also be true.

If that is so, then it can also be said that

f(n) = (1+2+3+...+(n-2))² + (n-1)³ + n³

is true, and

f(n) = (1+2+3+...+(n-3))² + (n-2)³ + (n-1)³ + n³

is true, and so on.

This process can be repeated until we reach a number which we have already proven - in this case, 1+2+3+4+5, which we can assert as true:-

(1+2+3+4+5)² = 1³+2³+3³+4³+5³

To be certain, though, we can take this to the extreme case, f(1):-

If g(2) is equal to h(1) + 2³

and h(1) = 1³

then g(1) must be 1²,

and since

1² = 1³

then g(2)=h(2) must be true, which means g(3)=h(3) must be true ... up to g(n)=h(n).

Therefore, for any arbitrary natural positive integer n,

I hope this proof meets with your approval. I submit it to the house.

998001 Redux

noreply@blogger.com (Alex Greene) — Tue, 31 Jan 2012 12:09:00 PST

I posted in an earlier article about how 1 / 998001 yields the decimal:-

0.000001002003004005006007008009 ... 995996997998999 ...

where the recurring element shows all of the three digit numbers, in sequence, from 000 to 999.

I commented on how 1 / 99980001 similarly yields:-

0.00000001000200030004000500060007 ... 999499959996999799989999

and 1 / 9801 yields:-

0.000102030405060708091011121314 ... 919293949596979899

I commented, further, how

9999 x 9999 = 99980001
999 x 999 = 998001
99 x 99 = 9801

What about the most trivial case, 9 x 9?

9 x 9 = 81

By this right, 1 / 81 should have a very specific pattern: the recurring portion has to be 0123456789 ...

A quick check confirms -

1 / 81 = 0.012345678901234567890123456789 ...

That would make 80 / 81

80 / 81 = 0.98765432109876543210 ...

Things get interesting when you look at 10/81:-

10/81 = 0.1234567890 ...

Just to confirm ...

(10 / 81) - (1 / 81) = 9 / 81 = 1 / 9

0.12345678901234567890 ...
0.01234567890123456789 ...
0.11111111111111111111 ...

which is, of course, the decimal expression of 1 / 9.

Wikipedia Article on Vedic Mathematics - FYI

noreply@blogger.com (Alex Greene) — Fri, 27 Jan 2012 01:46:00 PST

In doing my research, I noticed that Vedic Mathematics now has a Wikipedia article.

Bharati Krishna Tirtha's Vedic Mathematics on Wikipedia

Presented for your reference.

The Number 998001

noreply@blogger.com (Alex Greene) — Fri, 27 Jan 2012 01:08:00 PST

Recently, I showed some children in the library how to quickly multiply any three digit number by 999 using a very simple trick.

- First, subtract one from the number you wish to multiply by 999. Write down this number; it is the first half of the product.

So, for instance, if you were multiplying 568 by 999, the first thing you do is write down 567.

- Second, perform "All from 9 and the last from 10" on the number being multiplied by 999, and write this down next to the first three digits. That's your number.

So, again if you were multiplying 568 by 999, performing Nikhilam on 568 yields 432. Putting that next to 567 yields 567432 - which is 568 x 999.

The number 998001 is actually 999² = 999 x 999. (998001 is also the last square number before 1,000,000)

Again, working it out is easy once you know how to multiply three digit numbers by 999: subtract 1 from 999 to yield 998, and perform Nikhilam on 999 to yield 001.

998,001.

And now I can report on a delightful piece of information that I read about here:- namely, that the decimal fraction 1 / 998001 yields what the author calls "a surprising result" -

According to the author, 1 / 998001 yields every 3 digit number from 000 through to 999, in order. Similarly, 1 / 9801 yields a decimal comprising every 2 digit number from 00 through 99 in order.

Incidentally, you can also any two digit figure by 99 using the exact same method as outlined above for multiplication by 999, only it yields a four digit number.

So, for instance, if you multiply 56 x 99, subtract 1 from 56 to yield 55, and perform Nikhilam on 56 to yield 44, which in turn yields the result 5544 for 56 x 99.

And what is 9801, but 99 x 99.

Two questions arise as a result of these revelations.

1. Does the number recur after 999 in the fraction 1 / 998001 (i.e. the fraction eventually reaches the sequence "... 990991992993994995996997998999" and repeats from "000")? (The second half of the recurring portion would thus be a compliment of the first half, meaning that anyone computing 1 / 998001 would only need to calculate the first 1500 places after the decimal point) - and would the same result be true for 1 / 9081?

2. Is there a pattern here, which would hold true for 1 / 99980001 (9999 x 9999), 1 / 9999800001 (99999 x 99999), and so on? In other words, would we expect a recurring decimal part of 1 / 99980001 to begin "0.0000000100020003000400050006000700080009001000110012001300140015 ..." and end with "... 9990999199929993999499959996999799989999" before repeating? How about "0.00000000010000200003000040000500006000070000800009 ... 9999599996999979999899999" for the recurring part of the fraction 1 / 9999800001?

Happy New Year 2012

noreply@blogger.com (The Vedic Maths Forum India) — Sat, 31 Dec 2011 10:30:00 PST

New Year is the time to
unfold new horizons & realize new dreams,
to rediscover the strength & faith within u,
to rejoice in simple pleasures &
gear up 4 a new challenges.
Wishing u a truly fulfilling 2012

PM declares 2012 as 'National Mathematical Year'

noreply@blogger.com (The Vedic Maths Forum India) — Mon, 26 Dec 2011 11:14:00 PST

Chennai: Declaring 2012 as the 'National Mathematical year' as a tribute to maths wizard Srinivasa Ramanujan, Prime Minister Manmohan Singh on Monday voiced concern over the "badly inadequate" number of competent mathematicians in the country.

He also said that the perception that pursuit of mathematics does not lead to attractive career possibilities "must change."

"It is a matter of concern that for a country of our size, the number of competent mathematicians that we have is badly inadequate", he said at a function to here mark the 125th birth anniversary of Ramanujan.

Singh also declared December 22, the birthday of Ramanujan, as 'National Mathematics Day.'
Students have not pursued mathematics at advanced levels over more than three decades, which has resulted in a decline in quality of mathematics teachers at schools and colleges, Singh who is on a two-day visit to the state, told a galaxy of academics at Madras University.
"There is a general perception in our society that the pursuit of mathematics does not lead to attractive career possibilities. This perception must change. This perception may have been valid some years ago, but today there are many new career opportunities available to mathematics and the teaching perception itself has become much more attractive in recent years", Singh said.
The Prime Minister said the mathematical community has a duty to find out "ways and means" to address the shortage of top quality mathematicians and reach out to the public, especially in the modern context, where mathematics has tremendous influence on every kind of human endeavour.
Noting that the Central government has pursued a policy of encouraging scientific activities of diverse kinds, the Prime Minister said, "Given our traditions, we naturally attach special importance to mathematics...in many ways, mathematics can be regarded as the mother science".
He said Ramanujan overcame formidable difficulties to reach the pinnacle of greatness, illustrating the inadequacy of University evaluation system in the early decades of the last century, while at the same time showing the system displayed enough flexibility to take care of mavericks like him.
"There have been many reforms since those days but there would still be talent which would elude proper evaluation. Our institutions of higher learning must be sensitive to this problem."
"A genius like Ramanujan would shine bright even in the most adverse of circumstances, but we should be geared to encourage and nurture good talent which may not be of the same calibre as that of Ramanujan", Singh said.
Honouring Professor Robert Kanigel, who has written a biography of Ramanujan, Singh said this book has made Ramanujan well known to the public at large all over the world.
He said the country was proud of Ramanujan and Tamil Nadu has a special claim on him for he was a Tamilian.
"Along with CV Raman and Subramanyam Chandrashekhar (both Nobel laureates), he is among the three great men of science and mathematics that Tamil Nadu and India have given to the world of modern times", he said.

Special Xmas Offer on 8 DVD Set on Vedic Maths

noreply@blogger.com (The Vedic Maths Forum India) — Tue, 13 Dec 2011 12:56:00 PST

Greetings for the Festive Season!

Its that time of the year again when we all want to Gift
our near and dear ones and make them feel special.

What better way to do this than by the 8 DVD Set on High
Speed Vedic Mathematics which offers a revolutionary way to
speed up Math Calculations from Ancient India.

This 8 DVD set which has content of over 18 hours is
available to you now only for limited time at a Special Saving
of US $115. Which means this costs you now only US $ 75
instead of our earlier price of US $ 190.Plus you also get a Free
300 Page Book authored by renowned expert in India.

You can avail of this Special Offer by going to this link
and paying by your Credit Card.

https://www.paypal.com/cgi-bin/webscr?cmd=_s-xclick&hosted_button_id=R2REFFQ2FWD4S

All orders are shipped in 1 day's time and would reach you
by International Courier UBX, UPS,FEDEX or DHL In 3-5 days
time.

We hope that we are able to add to your Christmas Cheer by
this offer.

Wishing you loads of prosperity, abundance, success and a
radiant health.

Merry Christmas.

Thanking you
Gaurav Tekriwal
President
The Vedic Maths Forum India
www.vedicmathsindia.org

Global Mental Maths Methods

noreply@blogger.com (The Vedic Maths Forum India) — Fri, 09 Dec 2011 13:45:00 PST

Sharing a Presentation which I made on Global Mental Maths Methods. Let us know if you find this useful.

Ban on calculators to boost kids’ maths

noreply@blogger.com (The Vedic Maths Forum India) — Wed, 07 Dec 2011 09:35:00 PST

PUPILS are to be banned from using calculators in primary schools — in a bid to better their maths skills.

Schools minister Nick Gibb yesterday insisted the crackdown was vital to stop kids being maths dunces.

He said: "They shouldn't be reaching for a gadget every time they need to do a simple sum. Children can become too dependent on calculators."

The move will see a shake-up of the maths exam sat by all 11-year-olds. A section testing them on the use of calculators will be axed.

Mr Gibb said calculators were partly to blame for one in five primary kids failing to reach the expected level in maths this year.

He warned: "Without a solid grounding in arithmetic and early maths in primary school, children go on to struggle with basic maths skills throughout their school careers.

"It also means they leave school without the knowledge to complete everyday tasks in their adult lives."

Figures yesterday showed 17 million adults have maths skills no better than children as young as nine — up two million on a similar survey eight years ago.

Source: The Sun

BBC News

Divisibility Rules For Powers Of 2 Redux

noreply@blogger.com (Blogannath) — Sun, 13 Nov 2011 08:45:00 PST

So, last week, I thought I had come up with general rules for checking divisibility by powers of 2 that involved splitting the last few digits of the number into chunks of various lengths, multiplying them by various coefficients, adding them up and seeing if the sum is divisible by the power of 2. I also mentioned that I could not verify whether those rules were correct because I was not a theoretical mathematician or number theorist.

Actually, you don't need to be either of those to verify divisibility rules or derive new ones. All you need is to know the basics of modulo arithmetic. In fact, I had already used modulo arithmetic once before to explain why divisibility rules derived using osculation work. I had forgotten about that when I did my latest work on divisibility rules for powers of 2.

Fortunately, I remembered that work right after publishing my previous post. It was quite an eureka moment when that happened! I spent some time using modulo arithmetic to verify whether the rules I postulated actually work. Some of them do, some of them don't. And I also used modulo arithmetic to derive rules for divisibility by powers of 2. There are a large number of such rules, and I have now put together tables of such divisibility rules in the latest post on my blog.

The exercise has been a lot of fun for me. The sense of discovery you get when you figure out that you can build on something simple and basic, to explore something seemingly unconnected and much more complex, is priceless. If you have been intrigued by my forays into divisibility rules, I encourage you to visit my blog, and read my latest post. This time, you just have to take them and apply them. You don't have to try to verify them, find counterexamples or anything like that. Thank you, and good luck!

Multiplication using the Nikhilam Sutra -Vedic Maths

noreply@blogger.com (Uma) — Sat, 12 Nov 2011 17:30:00 PST

Vedic Maths ,as I have said in my last post is a boon for Math Lovers.Multiplication plays an important role from the elementary class to college to our day to day life.We know that without the 14 sutras ,Vedic Math wont exists.In fact,the fourteen sutras are the life line of Vedic system of Mathematics.Now,I am going to explain a simple process of Multiplication using one of the sutra named " The Nikhilam Sutra - All from 9 and last from 10".

Example 1:

605 x 504 = ?

The nearest base for the above two numbers are 500.So we fix this 500 as our base.Now,see that the above both numbers are greater than the base

How much deficient are are the above numbers are from base ?

605 -500 = +105
504 -500 = +004

Next,align as follows,

605 + 105
504 + 004
----------------
L.H.S 609 | ( + 420) R.H.S ---> [the RHS is got by adding diagonally on any one side ,the LHS is the result of multiplying the deficiencies .]
(609 /2) | (+420) since we have fixed base 500 and 500 is half of 1000 .i.e;1000/2 = 500.
304.5 + 420
304.5 | 420
Now see that the L.H.S has a decimal point ,so carry out that half to the R.H.S side from L.H.S .Now R..H.S side becomes 420 +500 =920

304 | 920
Therefore ,605 x 504 = 304920

Example 2

485 x 475 = ?

The nearest base for the above two numbers are 500.So we fix this 500 as our base.Now, see that the above both numbers are less than the base.

How much deficient are are the above numbers are from base ?

485 -500 = -015
475 -500 = -025

485 - 015
475- 025
--------------

460 | ( +375) -->the RHS is got by adding diagonally on any one side ,the LHS is the result of multiplying the deficiencies .

230 | 375 ---- (Half of 460 is 230)

230 | 375

Therefore 485 x 475 = 230375

Example 3

614 x 495 =?.

The nearest base for the above two numbers are 500.So we fix this 500 as our base.Now, see that the above numbers .One is greater than the base and one is less than the base.

How much deficient are are the above numbers are from base ?

614 - 500 = +114
495 - 500 = -005

614 + 114
495 - 005
------------

609 | (-570)--->-->the RHS is got by adding diagonally on any one side ,the LHS is the result of multiplying the deficiencies .
(609 /2) | (-570)
304.5 | (-570) ----- > the minus sign is removed by 1000 -570 =430 and remove one from the L.H.S side
303.5 | 430
303| (430+500) ----> the 0.5 when carried to R.H.S, becomes 500 and add it to R.H.S
303| 930

Therefore ,614 x 495= 303930

Note :The above three methods are general rule.When the multipliers are both above or greater we dont have any problems.But when one is greater and one is lesser,we follow the rules in example 3.For all ,the same rules apply.

The (0.5) on L.H.S is carried as 500 and added to the R.H.S every time you fall into a decimal (1/2).
The Minus signs resulting on R.H.S is removed by adding 1000 to it.

The theory behind the above rules is beyond my time here on this post.Kindly refer Bharati Krishna Tirtha Vedic Math

Arithmetic Mutiplications becomes very easy if you understand the sutras and the modus operandi involved in these sutras.Yet another popular method of multiplication is by Urdhva-Tiragbhyaam Sutra which uses 'Vertically and Crosswise' as its modus operandi.

Try these at home and put them in the comments below with the simple steps as above.
1. 510 x 610
2. 390 x 485
3. 460 x 680

About the Author :
Author Umamaheswari Anandane is a freelance writer , a poetess and a blogger at Inside My Poem Book and Perpetual Mind.You can visit her Blog link "Vedic Maths" for more of her writings on the same.

Divisibility Rules For Powers Of 2

noreply@blogger.com (Blogannath) — Sat, 05 Nov 2011 17:46:00 PDT

In an earlier post on the topic of divisibility rules, I posted some divisibility rules for 2, 4 and 8. I also posted a divisibility rule for 16 which was patterned after the ones for 2, 4 and 8. It turns out that this rule I had "derived" for 16 is completely wrong. And it is in fact, easy to verify that it is wrong, but for some reason, I neglected to verify it (and so far, nobody has pointed out to me that it is wrong using the comments on that post, so it seems as if nobody has bothered to verify it).

Even worse, I implied in that post that the divisibility rules for 2, 4, and 8 can be extended to any power of 2. Given that the extension does not even work for 16, it is obvious that this implication is not correct either.

In any case, revisiting these divisibility rules, and realizing that the rule I published for 16 was wrong, I started thinking about divisibility rules for powers of 2 once again. The simple rule that a number is divisible by 2^n if the last n digits are divisible by 2^n is easy to apply for 2 and 4, and perhaps even for 8. But for numbers like 16 and 32, application of the divisibility rule is almost as laborious as doing the division itself. As such, the main purpose of a divisibility rule, which is to verify whether division without a remainder is possible without actually performing the division, is compromised when you have to check divisibility of the last 5 digits of a number by 32, for instance. And it only gets worse for higher powers of 2.

Based on a lot of thinking and experimentation, I have now come up with what appear to be simpler divisibility rules for powers of 2 that have stood the test of some verification using real numbers (as opposed to verification using the logic that since they are simply an extension of a rule that works for some other numbers, they must be correct!). The rules are explained in this latest post on my blog, and I also show readers how to derive divisibility rules for any arbitrary powers of 2. These derived divisibility rules have the potential to be much simpler to apply than performing full-scale divisions of multi-digit numbers by powers of 2 to verify divisibility.

If you are interested, please visit my blog to read the entire post. This time, I have verified the newly derived rules with a few examples, but this is by no means exhaustive, and it is certainly no proof of the correctness of these rules. If you think you will find these rules useful (at least to stump colleagues or friends at get-togethers), please apply them to some examples and let me know if they work. If you use my derivation to extend these rules to higher powers of 2, let me know whether they work also.

I am not a theoretical mathematician or number theorist, so I don't know how to mathematically prove that these divisibility rules are correct. If I had been one of them, it probably would not have taken me quite so long to come up with these rules in the first place! But if one of my readers feel like taking on a challenge, please feel free to try to prove the correctness of these divisibility rules. And let me know if you find a proof (of either correctness or incorrectness). Thank you, and good luck!

Media: November 2011

noreply@blogger.com (The Vedic Maths Forum India) — Fri, 04 Nov 2011 06:55:00 PDT

The Magazine 'One India- One People' November Issue publishes an article on Vedic Maths- Vertically & Crosswise.The series of articles continue to be published. We would like to thank the editorial for the same.

Also a Bengali article got published about a workshop which was conducted in Rural Bengal, Birbhum district. We would like to thank Mr.Sitangshu Ghoshal, for taking great pains to organize the same.

The Geometry of Multiplication of Fractions

noreply@blogger.com (Guillermo Bautista) — Tue, 01 Nov 2011 20:15:00 PDT

Before algebra was invented, mathematicians in the early times represent mathematical expressions, equations, and proofs geometrically and verbally. In this post, we do the same: we explore a geometric representation of multiplication of fractions. I am not sure though if this strategy was used before.

We have discussed that the area of a rectangle is the product of its base and its height. A rectangle with base 5 units and height 3 units has area 15 square units.

This method can be extended to fractions. For example, how do we represent the multiplication of and ?

First, we represent the rectangle as a whole, then represent the fraction by dividing the rectangle horizontally into equal parts. Then, we shade the fractional part as shown. Two thirds is represented by two shaded rectangles out of three.

We do the same with our second fraction. We represent a whole with a rectangle with the same size and shape as that of the rectangle above. We then divide the rectangle vertically and shade three rectangles out of the five to represent three fifths.

To multiply the two fractions, we overlap the two rectangles. The number of squares formed is 15 and the number of shaded squares that intersect is 6.

Therefore,

Of course, instead of drawing two rectangles and overlapping, we can just draw one rectangle, and divide and shade them vertically and horizontally.

The Author

Guillermo Bautista is a mathematics teacher and a GeoGebra Institute Trainer in the Philippines. He is the writer of Mathematics and Multimedia and GeoGebra Applet Central.

An Ambassador of Vedic Maths

noreply@blogger.com (The Vedic Maths Forum India) — Sat, 29 Oct 2011 13:55:00 PDT

An Ambassador of Vedic Maths

This story titled 'An Ambassador of Vedic Maths' just got published in the Good News Chronicle.

For full story go here http://thegoodnewschronicle.com/ts201110151.asp

Happy Diwali

noreply@blogger.com (The Vedic Maths Forum India) — Tue, 25 Oct 2011 09:55:00 PDT

Update TIME 2011 - Math Carnival - Ganit Utsav

noreply@blogger.com (The Vedic Maths Forum India) — Mon, 17 Oct 2011 05:21:00 PDT

Math Carnival at TIME 2011

View more documents from gtekriwal

82nd Carnival of Mathematics

noreply@blogger.com (The Vedic Maths Forum India) — Fri, 14 Oct 2011 12:49:00 PDT

Welcome to the 82nd Carnival of Mathematics. As the tradition goes we begin by stating some trivia about the number 82.

82 is a Happy Number .It is also the 23rd biprime and if you consider physics for a while , 82 is the sixth Magic number.

Magic,Maths and Happiness reminds me of the great Martin Gardner and his legacy.This coming October 21, 2011, will see the second annual MartinGardner Worldwide Celebration of Mind. Last year it had over sixty events around the globe to celebrate Martin's birthday and his legacy in recreational mathematics. People gathered in informal groups to share puzzles, mathematical games, magic, and many other topics that Martin loved so much.

We begin by sharing Earl Samuelson's post on The Mathematics of Music on his blog. There is indeed a massive amount of mathematics occurring within music. It is present in the form of ratios,frequency modulations etc.

Guillermo Bautista in his blog Mathematics and Multimedia explains to us what PowerSets are with his post titled Milkshakes and Powersets and NerdMom at Frugal HomeSchooler shares various websites which offer help in Fractions.

The Institute of Mathematics and Computer Science IMACS give us an insight on how to keep talented girls on the STEM track.

Feanor's post titled Divisive Diversions talks about divisibility problems and patterns from computing while John Cook presents Willie Sutton and the multivariate normal distribution — The Endeavour posted at The Endeavour.

Ron Doerfler and Miles Forster from the Dead Reckonings Blog in his 2-part essay titled The 13th Root of a 100-Digit Number shows us a historical overview of the extraction of 13th roots, including the methods used by a few mental calculators, methods that largely rely on a mix of intensive mental calculation and large-scale rote memorization. It demonstrates the creativity and drive of these marvelous people.

Gianluigi Filippelli in his blog Doc Madhattan presents to us the story of the E8 group in the post the Universe and the flowers.

João F. Ferreira shares with us An improved proof of the handshaking lemma on his blog João F. Ferreira | Programming, Algorithms, and Calculational Mathematics. Brent Yorgey presents Some words about Post without words #2 posted at The Math Less Traveled.

And lastly Alex Greene at the Vedic Maths Forum India Blog gives in detail, the Progression of Squares and Cube Numbers.

Progressions of Square and Cube Numbers

noreply@blogger.com (Alex Greene) — Tue, 11 Oct 2011 12:10:00 PDT

Square Numbers

Some of you may have studied the progression of square numbers, and discovered the rather elegant pattern which connects them - namely, that between two adjacent square numbers, the difference is always going to be an odd number; and with each step to the next square number, the difference always increases. This has always been a useful guide in predicting when each next square number should be; but what is the formula by which you can work out what that next square number is?

Well, let's look at the smallest square numbers - the square numbers for 1², 2², 3², 4² and 5².

Here are the first five square numbers, and for completion let's throw in 0² as well:-

Number	Square	Difference
0²	0	N/A
1²	1	1
2²	4	3
3²	9	5
4²	16	7
5²	25	9

As you can see, not only is the difference an odd number each time; it is a unique odd number. Other than 1 and 4, no other pair of square numbers will ever be found that will differ by 3; beside 25 and 36, no other pairing of square numbers will ever have a difference of 11.

The very specific progression between square numbers reveals a pattern, which can be formulated. But what is the formula for this difference between the squares?

Let's look at a pair of square numbers; 1 and 4, 1² and 2². The difference, 3, can be considered (2x1) + 1. Look at the difference between 16 and 25, or 4² and 5² respectively. 9 = (2x4) + 1.

Now let's abstract this to the case of the number n, where

Q_n = n²

and

Q_n+1 = (n + 1)².

The difference between Q_n and Q_n+1 is given as:-

Q_n+1 - Q_n = (n + 1)² - n²

= (n + 1)(n + 1) - n²

= n² + 2n + 1 - n²

Q_n+1 - Q_n = 2n + 1.

Let's take this formula to look at the difference between two adjacent squares for an arbitrary value, say 37². What is the next square number, 38², going to be?

Using the formula 2n + 1, and substituting n = 37, we arrive at:-

(2 x 37) + 1 = 74 + 1 = 75.

A quick calculation, and I work out that 37² = 1369. Adding 75 to that, we get 1444 - and the square root of 1444 is ... 38.

One last test, just in case; two adjacent large numbers, 13365 and 13366.

13365² = 178,623,225

(2 x 13365) + 1 = 26731

178623225 + 26731 = 178649956

sqrt(178649956) = 13366.

So there you have it. If you know one square number n² and its corresponding square root n, you can work out its adjacent neighbour by calculating the difference, 2n + 1.

But there's more. If you know the number n, you can work out the neighbour to n + 1, n + 2, n + 3 ... and also go back, subtracting 2(n-1) + 1, 2(n-2) + 1, 2(n-3) + 1 ..., always using the same formula, each time. Handy to know, say, if you were challenged to calculate not only, say, 999² (998,001 of course), but also the next ten square numbers above and below it ...

Cube Numbers

Does a similar relationship exist between adjacent cube numbers - between, say, 4³ and 5³, or between 1741³ and 1742³, or between n³ and (n + 1)³?

Let's see.

Number	Square	Difference
0³	0	N/A
1³	1	1
2³	8	7
3³	27	19
4³	64	37
5³	125	61

This pattern is not so easy to spot, so let's look at the difference R between R_n and R_n+1 where

R_n = n³

and

R_n+1 = (n + 1)³

R = R_n+1 - R_n

= (n + 1)³ - n³

= (n + 1)(n + 1)(n + 1) - n³

= n³ + 3n² + 3n + 1 - n³

= 3n² + 3n + 1

= 3n(n + 1) + 1.

R = 3n(n + 1) + 1.

Can we confirm that this formula works? Let's try three examples, namely 5³ and 6³, 74³ and 75³, and 1741³ and 1742³.

5³ = 125.

R = 3 x 5 (5 + 1) + 1

R = (15 x 6) + 1

R = 91

125 + R = 216 = 6³.

Next, let's look at 74³ and 75³;

74³ = 405224

R = (3 x 74 x 75) + 1

R = (222 x 75) + 1

R = 16651

125³ + R = 405224 + 16651

= 421875

= 75³.

And now the last one, just to make sure:-

1741³ = 5,277,112,021

R = ((1741 x 3) x 1742) + 1

= (5223 x 1742) + 1

= 9098466 + 1

= 9098467

5277112021 + 9098467 = 5286210488 = 1742³.

Pretty conclusive, then.

Conclusion

So, then, it's pretty clear that to determine the neighbouring number for any given square number N = n², you can use the formula

2n + 1

and to determine the neighbour of any cube number N = n³, you can use the formula

3n(n + 1) +1.

I leave the determination of the formulae to calculate the differences between adjacent integers of higher powers to the reader. One hint I will leave you with: look to Pascal's Triangle.

82nd Carnival of Mathematics

noreply@blogger.com (The Vedic Maths Forum India) — Tue, 11 Oct 2011 11:55:00 PDT

We are proud to be hosting the 82nd Carnival of Mathematics on the 14th October 2011.We had earlier hosted the 12th Carnival of Mathematics back few years ago and it feels good to be hosting it again. So I would like to invite everybody to contribute and make it a grand sucesss.

You can use this form for your submissions.

http://blogcarnival.com/bc/submit_1049.html

The Nine Digit Squares Puzzle - Answers

noreply@blogger.com (Alex Greene) — Sat, 08 Oct 2011 13:38:00 PDT

Here is the answer to the puzzle I posted a few days back. The original puzzle was:-

Nine digits, other than zero, can be arranged uniquely to form four numbers which are all squares:-

9 (3²) 81 (9²) 324 (18²) 576 (24²)

What is the smallest possible square number comprising all nine digits together, and the largest square number comprising all nine digits together?

The answers are:- (smallest) 139,854,276 (11826²) and (largest) 923,187,456 (30384²).

The first correct answer came in from Balaji Ramanathan. Congratulations.

The Puzzle - Answers Tomorrow

noreply@blogger.com (Alex Greene) — Fri, 07 Oct 2011 16:20:00 PDT

I have not forgotten my duty to provide the answer to the puzzle I set some days ago. I have received a correct answer already, and credit will be given - I just needed to give time to everyone here to come up with answers for themselves.

I will provide the answer to the puzzle on Saturday, 8 October, in the evening (GMT).