## Sunday, March 27, 2011

### Consecutive numbers and consecutive prime factors

http://www.primepuzzles.net/puzzles/puzz_240.htm

Example:

384 = 2^7 * 3 .................................... 385 = 5 * 7 * 11

539 = 7^2 * 11 ................................... 540 = 2^2 * 3^3 * 5

### Montessori : Education, Philosophy and Experiences

Google Founders discuss their experience as Montessori students

### Proof that 22/7 exceeds π

## Saturday, March 26, 2011

### Conjecture about sum of primes

1 + 2 + 2^2 + 2^3 + 2^4 = 1 + 5 + 5^2 = 31

(2) research with .... a + p + p^2 + p^3 + .... = a + ....

### Squares which remain squares if you increment every digit by 1

http://oeis.org/A061843

0

25

2025

13225

4862025

60415182025

207612366025

153668543313582025

13876266042653742025

20761288044852366025

47285734107144405625

406066810454367265225

141704161680410868660551655625

## Friday, March 25, 2011

### Factorials

0! = 1

1! = 1

2! = 2

3! = 6

4! = 24

5! = 120

6! = 720

7! = 5040

8! = 40320

9! = 362880

10! = 3628800

11! = 39916800 ........ 66 weeks

12! = 479001600 ....... 792 weeks (792 / 66 = 12 )

13! = 6227020800 ...... 10296 weeks (10296 / 792 = 13)

14! = 87178291200 ...... 144144 weeks (144144 / 10296 = 14 )

15! = 1307674368000

16! = 20922789888000

17! = 355687428096000

18! = 6402373705728000

19! = 121645100408832000

20! = 2432902008176640000

21! = 51090942171709440000

1 hour = 3600 seconds

1 day = 86400 seconds

1 week = 604800 seconds .... 6 weeks = 604800 * 6 = 3628800 = 10!

1 year = 31556926 seconds ... 11! is larger than the number of seconds in a year

[ 39916800 - 31556926 = 8359874 seconds or 96.75780092592592 days]

1! = 1

2! = 2

3! = 6

4! = 24

5! = 120

6! = 720

7! = 5040

8! = 40320

9! = 362880

10! = 3628800

11! = 39916800 ........ 66 weeks

12! = 479001600 ....... 792 weeks (792 / 66 = 12 )

13! = 6227020800 ...... 10296 weeks (10296 / 792 = 13)

14! = 87178291200 ...... 144144 weeks (144144 / 10296 = 14 )

15! = 1307674368000

16! = 20922789888000

17! = 355687428096000

18! = 6402373705728000

19! = 121645100408832000

20! = 2432902008176640000

21! = 51090942171709440000

1 hour = 3600 seconds

1 day = 86400 seconds

1 week = 604800 seconds .... 6 weeks = 604800 * 6 = 3628800 = 10!

1 year = 31556926 seconds ... 11! is larger than the number of seconds in a year

[ 39916800 - 31556926 = 8359874 seconds or 96.75780092592592 days]

31556926 * 12 = 378683112

12! is larger than the number of seconds in 12 years

13! is larger than the number of seconds in a century.

5! - 4! = 96 seconds

6! - 5! = 600 seconds

7! - 6! = 4320 seconds = 0.05 Days

8! - 7! = 35280 seconds = 0.4083333333333333 Days

9! - 8! = 322560 seconds = 3.7333333333333334 Days = 0.533333333 week

10! - 9! = 3265920 seconds = = 37.8 Days = 5.4 weeks

11! - 10! = 36288000 seconds = 420 days = 60 weeks

12! is larger than the number of seconds in 12 years

13! is larger than the number of seconds in a century.

5! - 4! = 96 seconds

6! - 5! = 600 seconds

7! - 6! = 4320 seconds = 0.05 Days

8! - 7! = 35280 seconds = 0.4083333333333333 Days

9! - 8! = 322560 seconds = 3.7333333333333334 Days = 0.533333333 week

10! - 9! = 3265920 seconds = = 37.8 Days = 5.4 weeks

11! - 10! = 36288000 seconds = 420 days = 60 weeks

1! + 2! = 3

1! + 2! + 3! = 9

1! + 2! + 3! + 4! = 33

1! + 2! + 3! + 4! + 5! = 153

1! + 2! + 3! + 4! + 5! + 6! = 873

1! + 2! + 3! + 4! + 5! + 6! + 7! = 5913

1! + 2! + 3! + 4! + 5! + 6! + 7! + 8! = 46233

1! + 2! + 3! + 4! + 5! + 6! + 7! + 8! + 9! = 409113 (0.676443452 week)

1! + 2! + 3! + 4! + 5! + 6! + 7! + 8! + 9! + 10! = 4037913 (6.676443452 weeks)

### Find all primes less than 100

that can each be written in all three ways:

p = a^2 + b^2 = c^2 + 2 d^2 = e^2 + 3 f^2

### Math Education: An Introduction to the Montessori Math Curriculum

An overview of how the Montessori approach to mathematics moves from the concrete to the abstract. The DVD covers the curriculum from age 3 through 12 and is available at www.edvid.com. It is 25 minutes in length.

## Thursday, March 24, 2011

### Amicable Numbers

220, 284

220 + 284 = 504 = 2^3 * 3^2 * 7

1184, 1210 (discovered by a 16-year-old Italian named Nicolo Paganini)

1184 + 1210 = 2394 = 2 * 3^2 * 7 * 19

17296, 18416

17296 + 18416 = 35712 = 2^7 * 3^2 * 31

9363584, 9437056

9363584 + 9437056 = 18800640 = 2^13 * 3^3 * 5 * 17

666030256, 696630544 whose sum is not divisible by 9 is Poulet's pair

666030256 + 696630544 = 1362660800 = 2^16 * 5^2 * 31 * 83 * 331

220 + 284 = 504 = 2^3 * 3^2 * 7

1184, 1210 (discovered by a 16-year-old Italian named Nicolo Paganini)

1184 + 1210 = 2394 = 2 * 3^2 * 7 * 19

17296, 18416

17296 + 18416 = 35712 = 2^7 * 3^2 * 31

9363584, 9437056

9363584 + 9437056 = 18800640 = 2^13 * 3^3 * 5 * 17

666030256, 696630544 whose sum is not divisible by 9 is Poulet's pair

666030256 + 696630544 = 1362660800 = 2^16 * 5^2 * 31 * 83 * 331

### Train Your Brain Like a Champ

Memory Champion Joshua Foer Reveals Secrets of the Brain

Building a Memory Palace Isn't Just a Parlor Trick, It Can Help in Daily Life

http://abcnews.go.com/US/memory-champion-joshua-foer-reveals-secrets-brain/story?id=13205718

The Memory Champion

2008 USA National Memory Champion, Chester Santos, prepares to defend his title.

Leaner Body Makes for Better Memory

A study finds that verbal memory increases with weight loss.

Healthy Life.

Your Diet Can Improve Your Memory.

Multitasking May Harm Memory

Also, Men With MS May Be More Likely to Pass It On

Amazing Memory Man Brain Being Tested By Scientists

Scientists Are Studying Brad Williams' Superior Memory

For more information about Brad Williams, e-mail unforgettabledoc@sbcglobal.net

The Earliest Fetal Memory?

Dutch Doctors Say the Unborn may Have Memories by the 30th Week of Pregnancy

Jill Price: The Woman Who Can't Forget

Building a Memory Palace Isn't Just a Parlor Trick, It Can Help in Daily Life

http://abcnews.go.com/US/memory-champion-joshua-foer-reveals-secrets-brain/story?id=13205718

The Memory Champion

2008 USA National Memory Champion, Chester Santos, prepares to defend his title.

Leaner Body Makes for Better Memory

A study finds that verbal memory increases with weight loss.

Healthy Life.

Your Diet Can Improve Your Memory.

Multitasking May Harm Memory

Also, Men With MS May Be More Likely to Pass It On

Scientists Are Studying Brad Williams' Superior Memory

The Earliest Fetal Memory?

Dutch Doctors Say the Unborn may Have Memories by the 30th Week of Pregnancy

## Wednesday, March 23, 2011

### English alphabet --> prime numbers (Recreational Maths)

English alphabet (26 letters) : A B C D E F G H I J K L M N O P Q R S T U V W X Y Z

The First 26 prime numbers: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101

A=2 ... B=3 ... C=5 ... D=7 ... E=11 ... F=13 ... G=17 ... H=19 ... I=23 ... J=29

K=31 ... L=37 ... M=41 ... N=43 ... O=47 ... P=53 ... Q=59 ... R=61

S=67 ... T=71 ... U=73 ... V=79 ... W=83 ... X=89 ... Y=97 ... Z=101

I'm looking for interesting patterns.

For example, take a 3-letter word, say CAR. We can either add or multiply their values.

Palindromic Words, Phrases and Sentences

The First 26 prime numbers: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101

A=2 ... B=3 ... C=5 ... D=7 ... E=11 ... F=13 ... G=17 ... H=19 ... I=23 ... J=29

K=31 ... L=37 ... M=41 ... N=43 ... O=47 ... P=53 ... Q=59 ... R=61

S=67 ... T=71 ... U=73 ... V=79 ... W=83 ... X=89 ... Y=97 ... Z=101

I'm looking for interesting patterns.

For example, take a 3-letter word, say CAR. We can either add or multiply their values.

If adding, we get ... CAR ---> 5 + 2 + 61 = 68

If we multiply, we get ... CAR ---> 5 * 2 * 61 = 610

(1) We could look for English words that are prime numbers, or squares or cubes, or any other interesting numbers.

(2) We could look at palindromic words and palindromic set of words (or phrases)

(3) We could look for English words that are palindromic

Palindromic Words, Phrases and Sentences

### Game Theory

How to Win Friends & Influence People

NYU’s Brams Applies Game Theory to the Humanities in New Book

Prof. Dr. Robert John Aumann on game theory, incentives and the financial crisis

Game Theory and Mutual Misunderstanding ebook

Harry Potter teaches game theory!

Game Theory and Mutual Misunderstanding ebook

## Sunday, March 20, 2011

### 12-sided die

Instead of having 1, 2, ..., 12, on the faces of the die, a 12-sided die is created by engraving the numbers 1- 6 on the faces of a regular dodecahedron so that…

(1) Each number appears twice, and

(2) Each occurrence of each number is surrounded by all five other numbers.

In how many ways can this be done?

Two ways are considered the same if one can be rotated to become the other

Bonus question:

What shapes make the best dice,

(a) tetrahedron,

(b) hexahedron,

(c) octahedron,

(d) dodecahedron,

(e) icosahedron,

(f) other?

### Circular Lake

A friend of yours is at the centre of a circular lake that has a radius of 1km.

She can row at 2km/hr in any sort of continuous path. Alas, you are on the shore of the lake but have no boat.

How fast must you be able to run so that you can be assured of catching her when she reaches the shore?

She can row at 2km/hr in any sort of continuous path. Alas, you are on the shore of the lake but have no boat.

How fast must you be able to run so that you can be assured of catching her when she reaches the shore?

## Friday, March 18, 2011

### Some Pi to Celebrate St. Patrick's Day!

This four-leaf clover consists of four coplanar circles. Each circle is externally tangent to two others, as shown. The two smaller circles are congruent, and the two larger circles are congruent. A square is constructed such that each of its four vertices is also the center of one of the four circles. Suppose the area of the square is 6 1/4 in2, and the combined area of the four circular leaves of the clover is 6 1/2π in2. What is the positive difference between the radius of a large circular leaf and the radius of a small circular leaf? Express your answer as a common fraction.

Another four-leaf clover also consists of 4 coplanar circles. The large circular leaves are externally tangent to each other, as well as to each of the smaller circular leaves, which are also congruent to one another. The radius of the large and small circular leaves is 1 1/2 inches and 1 inch, respectively. What is the area of a rhombus formed such that each of its vertices is also the center of one of the four circular leaves?

Three-leaved clover

### Education: Russian Math

Web-Based Russian Math Curriculum Shows Positive Results

Mathematical Circles and Olympiads

Russia's Conquering Zeros

The strength of post-Soviet math stems from decades of lonely productivity

The strength of post-Soviet math stems from decades of lonely productivity

About the Independent University of Moscow

Critical Issues in Education: Teaching Teachers Mathematics

## Tuesday, March 15, 2011

### The Way You Learned Math Is So Old School

http://www.npr.org/2011/03/05/134277079/the-way-you-learned-math-is-so-old-school?sc=tw

A stemplot (or stem-and-leaf plot)

## Monday, March 14, 2011

### Math quotes #3

In engineering there is an adage: “if you are one step ahead, you are a genius; if you are two steps ahead, you are an idiot!”

@mathematicsprof added: and if you're 3 steps ahead, you're probably a religious prophet

The more you study logic, the more you value coincidence.

(from the movie Fermat's Room)

@mathematicsprof added: and if you're 3 steps ahead, you're probably a religious prophet

The more you study logic, the more you value coincidence.

(from the movie Fermat's Room)

## Saturday, March 12, 2011

### Math Reading List

Mathematicians: An Outer View of the Inner World

Book Review

Euler's Gem: The Polyhedron Formula and the Birth of Topology

Book Review

Gamma: Exploring Euler's Constant

Book Review

The Road to Reality: A Complete Guide to the Laws of the Universe

Book Review

The Math Forum @ Drexel: Math Books Recommended

Book Review

Euler's Gem: The Polyhedron Formula and the Birth of Topology

Book Review

Gamma: Exploring Euler's Constant

Book Review

The Road to Reality: A Complete Guide to the Laws of the Universe

Book Review

The Math Forum @ Drexel: Math Books Recommended

### Fermat Corner

A great read if you want to get some insights to the mind of a mathematician :

http://www.simonsingh.net/Fermat_Corner.html

http://www.simonsingh.net/Fermat_Corner.html

## Wednesday, March 9, 2011

### John Von Neumann Interview

John Von Neumann appears on the television program "America's Youth Wants To Know". He made this appearance when he was the Commissioner of the Atomic Energy Commission.

### I Want to Be a Mathematician: A conversation with Paul Halmos

The 44-minute film contains a rare interview with Paul Halmos by Peter Renz, revealing his thoughts on mathematics, and how to teach it and write about it. Five bonus features include comments by mathematicians Robert Bekes, David Eisenbud, Jean Pedersen, and Donald Sarason about their experiences with Halmos. Interviews with Halmos by Don Albers and Halmos's own writings are included as PDF documents.

### Hilbert's Hotel and Infinity

http://www.reasonablefaith.org/site/PageServer

Dr. William Lane Craig proves Al Ghazali's premise that an actual infinite number of things is absurd by using Hilbert's Hotel (if you're an atheist listen carefully: Dr. Craig did not say actual infinites are non-existent). The following clip comes from Dr. Craig's lecture

This is part of the "Defending the Cosmological Argument" series. Table of Contents:

http://www.youtube.com/view_play_list?p=916E17EE70E98A68

http://www.youtube.com/view_play_list?p=916E17EE70E98A68

Dr. William Lane Craig proves Al Ghazali's premise that an actual infinite number of things is absurd by using Hilbert's Hotel (if you're an atheist listen carefully: Dr. Craig did not say actual infinites are non-existent). The following clip comes from Dr. Craig's lecture

### Julia Robinson and Hilbert's Tenth Problem

Julia Robinson and Hilbert's Tenth Problem features a heroine driven by the quest to solve one of the central problems of modern mathematics. She rises above formidable obstacles to assume a leading role in her field. Julia Robinson was the first woman elected to the mathematical section of the National Academy of Sciences, and the first woman to become president of the American Mathematical Society. While tracing Robinson's contribution to the solution of Hilbert's tenth problem, the film illuminates how her work led to an unusual friendship between Russian and American colleagues at the height of the Cold War.

### World mental arithmetic championships

What is 6104179 divided by 6913? Can you work it out in your head? And can you do it in under 12 seconds? If so, you could compete at the world mental arithmetic championships, which we visited to meet some of top up-and-coming number crunchers.

## Tuesday, March 8, 2011

### More applications of winding numbers

We define the degree of a function from the circle to the circle, and use that to show that there is no retraction from the disk to the circle, the Brouwer fixed point theorem, and a Lemma of Borsuk.

### Introduction to Algebraic Topology

This is the first video of the introductory lecture of a beginner's course in Algebraic Topology. The subject is one of the most dynamic and exciting areas of 20th century mathematics, with its roots in the work of Riemann, Klein and Poincare in the latter half of the 19th century. This first lecture will outline the main topics, and will present three well-known but perhaps challenging problems for you to try.

The course is for 3rd or 4th year undergraduate math students, but anyone with some mathematical maturity and a little background or willingness to learn group theory can benefit. The subject is particularly important for modern physics. Our treatment will have many standard features, but also some novelties.

The lecturer is Assoc Prof N J Wildberger of the School of Mathematics and Statistics at UNSW, Sydney, Australia.

The course is for 3rd or 4th year undergraduate math students, but anyone with some mathematical maturity and a little background or willingness to learn group theory can benefit. The subject is particularly important for modern physics. Our treatment will have many standard features, but also some novelties.

The lecturer is Assoc Prof N J Wildberger of the School of Mathematics and Statistics at UNSW, Sydney, Australia.

### The Ham Sandwich theorem and the continuum

In this video we give the Borsuk Ulam theorem: a continuous map from the sphere to the plane takes equal values for some pair of antipodal points. This is then used to prove the Ham Sandwich theorem (you can slice a sandwich with three parts (bread, ham, bread) with a straight planar cut s so that each slice is cut cut in two. Also we give an application to the continuum: the plane is different (not homeomorphic) 3 dimensional space.

### Look-and-say sequence

It is the sequence of integers beginning as follows:

1, 11, 21, 1211, 111221, 312211, 13112221, 1113213211, ...

1, 11, 21, 1211, 111221, 312211, 13112221, 1113213211, ...

### Doomsday algorithm

This little batch calculates the weekday of any date between 01.01.0000 and 31.12.9999. The algorithm is from john horton conway.

## Monday, March 7, 2011

### THE PITCH DROP EXPERIMENT

John Mainstone and the late Thomas Parnell of the University of Queensland, Australia, for patiently conducting an experiment that began in the year 1927 -- in which a glob of congealed black tar has been slowly, slowly dripping through a funnel, at a rate of approximately one drop every nine years.

http://www.physics.uq.edu.au/physics_museum/pitchdrop.shtml

http://www.physics.uq.edu.au/physics_museum/pitchdrop.shtml

### Spontaneous knotting of an agitated string

Dorian Raymer of the Ocean Observatories Initiative at Scripps Institution of Oceanography, USA, and Douglas Smith of the University of California, San Diego, USA, for proving mathematically that heaps of string or hair or almost anything else will inevitably tangle themselves up in knots.

### Preventing winter falls:

A randomised controlled trial of a novel intervention

Lianne Parkin, Sheila Williams, and Patricia Priest of the University of Otago, New Zealand, for demonstrating that, on icy footpaths in wintertime, people slip and fall less often if they wear socks on the outside of their shoes.

### why spaghetti do not break in half

why, when you bend dry spaghetti, it often breaks into more than two pieces.

### Scientists say the darnest thing

Isaac Asimov said: ‘The most exciting phrase to hear in science is not 'Eureka!'

What is the greatest mathematical discovery of all time?

but 'That's funny!'

What is the greatest mathematical discovery of all time?

Compound Interest according to Einstein!!

### Kissing Number

Kissing number problem

...from the description of a talk to be given on Monday, March 7, 2011 at Northeastern University, by mathematician Abhinav Kumar (whose Erdös Number is 2 — a fact of interest to those interested in Erdös Numbers.

List of people by Erdős number

MathSciNet: Collaboration distance - American Mathematical Society

The Erdös Number Project http://www.oakland.edu/enp/

## Sunday, March 6, 2011

### Of Conjectures

... And BIG NUMBERS

Funny comix:

Clay Mathematics Institute: We give up

Search for counter examples to conjectures

Mersenne conjectures

Euler observed that the Mersenne number at M_61 is prime, so refuting the conjecture.

M_61 is a very large number.

Euler's sum of powers conjecture

Check the counterexamples for k=4 and k=5

Chebyshev Bias: Not very large, but large enough

**Archimedes' cattle problem : A large number**

Goldbach's Conjecture verified results: verified up to 10^18

No counter example found.

Pólya conjecture

An explicit counterexample, of n = 906,180,359 was given by R. Sherman Lehman in 1960. The smallest counterexample is n = 906,150,257, found by Minoru Tanaka in 1980

Big Number: The sum of Scrooge's wealth is very controversial

no actual counter-example has been found.

Mertens conjecture

Other examples can be found here

## Saturday, March 5, 2011

### How to use geometry for propositional betting

Learn How to use geometry for propositional betting. For more Bar Tricks How-To Videos & Articles, visit WonderHowTo.

How to force the outcome of a coin toss with a simple trick

Learn How to force the outcome of a coin toss with a simple trick. For more Cons How-To Videos & Articles, visit WonderHowTo.

Learn How to predict the outcome of a coin toss when you flip. For more Prop Tricks How-To Videos & Articles, visit WonderHowTo.

Learn How to do a cool coin flip switch magic trick. For more Prop Tricks How-To Videos & Articles, visit WonderHowTo.

### Pi Day

József Kürschák

Albert Einstein

Waclaw Sierpinski

Since Sierpinski (ala Sierpinski triangle) was born on Pi day we should make a Sierpinski pie. Here's the recipe

Nora Isobel Calderwood

Nora Isobel Calderwood

Translations of pi

After his death the "Ludolphine number", 3.14159265358979323846264338327950288...,

was engraved on his tombstone

The tombstone of Ludolph van Ceulen

50 Interesting Facts About Pi

How many different ways to express pi ? Well here are 59 for a starter

What Pi sounds like

Ramanujan proved the following formula:

http://planetmath.org/encyclopedia/RamanujansFormulaForPi.html

Pi and the Fibonacci Numbers

Anti-Pi

### Factorials and Trailing Zeroes

23! has four trailing zeroes

101! has 24 trailing zeroes

1000! has 249 trailing zeroes

4617! has 1151 trailing zeroes.Note that 41! = 33452526613163807108170062053440751665152000000000

It is a 50 digits long, the last 9 of them are 0s. So the trailing zeros represent 18%

Find n where n! has the largest possible proportion of trailing 0s.

101! has 24 trailing zeroes

1000! has 249 trailing zeroes

4617! has 1151 trailing zeroes.

It is a 50 digits long, the last 9 of them are 0s. So the trailing zeros represent 18%

Find n where n! has the largest possible proportion of trailing 0s.

## Friday, March 4, 2011

### Number Patterns #2

1 * 8 + 1 = 9

12 * 8 + 2 = 98

123 * 8 + 3 = 987

1234 * 8 + 4 = 9876

12345 * 8 + 5 = 98765

123456 * 8 + 6 = 987654

1234567 * 8 + 7 = 9876543

12345678 * 8 + 8 = 98765432

123456789 * 8 + 9 = 987654321

### Number Patterns

My aim is to find sets of integers, A and B, such that

(1) A = (a1, a2, ..., ai) and B = (b1, b2, ..., bj)

(a1) + (a2) + ... + (ai) = (b1) + (b2) + ... + (bj)

(a1)^2 + (a2)^2 + ... + (ai)^2 = (b1)^2 + (b2)^2 + ... + (bj)^2

(a1)^3 + (a2)^3 + ... + (ai)^3 = (b1)^3 + (b2)^3 + ... + (bj)^3

(a1)^4 + (a2)^4 + ... + (ai)^4 = (b1)^4 + (b2)^4 + ... + (bj)^4

(a1)^5 + (a2)^5 + ... + (ai)^5 = (b1)^5 + (b2)^5 + ... + (bj)^5

and,

(a1 + 1)^2 + (a2 + 1)^2 + ... + (ai + 1)^2 = (b1 + 1)^2 + (b2 + 1)^2 + ... + (bj + 1)^2

(a1 + 2)^2 + (a2 + 2)^2 + ... + (ai + 2)^2 = (b1 + 2)^2 + (b2 + 2)^2 + ... + (bj + 2)^2

(a1 + 3)^2 + (a2 + 3)^2 + ... + (ai + 3)^2 = (b1 + 3)^2 + (b2 + 3)^2 + ... + (bj + 3)^2

(a1 + 4)^2 + (a2 + 4)^2 + ... + (ai + 4)^2 = (b1 + 4)^2 + (b2 + 4)^2 + ... + (bj + 4)^2

(2) A = (a1, a2, ..., ai) and B = (b1, b2, ..., bj)

(a1 + p)^2 + (a2 + p)^2 + ... + (ai + p)^2 = (b1 + p)^2 + (b2 + p)^2 + ... + (bj + p)^2

p being the first prime numbers 2, 3, 5, 7

Here are set of equations in which the sums of powers of two different sets of numbers are the same for several different exponents. The simplest example :

The two sets, here, are: (1, 6, 8) and (2, 4, 9)

Remarkably, if any integer is added to all terms of the equation, it will still hold

Adding 1 to the example above, gives

Adding 2 and 3 to all terms of the equation 1 + 8 + 10 + 17 = 36 = 2 + 5 + 13 + 16

(1+2)^2 + (8+2)^2 + (10+2)^2 + (17+2)^2 = 614

(2+2)^2 + (5+2)^2 + (13+2)^2 + (16+2)^2 = 614

(1+3)^2 + (8+3)^2 + (10+3)^2 + (17+3)^2 = 706

(2+3)^2 + (5+3)^2 + (13+3)^2 + (16+3)^2 = 706

(1+4)^2 + (8+4)^2 + (10+4)^2 + (17+4)^2 = 806

(2+4)^2 + (5+4)^2 + (13+4)^2 + (16+4)^2 = 806

OR

AND

1 + 50 + 57 + 15 + 22 + 71 = 216 = 2 + 45 + 61 + 11 + 27 + 70

1^2 + 50^2 + 57^2 + 15^2 + 22^2 + 71^2 = 11500

2^2 + 45^2 + 61^2 + 11^2 + 27^2 + 70^2 = 11500

1^3 + 50^3 + 57^3 + 15^3 + 22^3 + 71^3 = 682128

2^3 + 45^3 + 61^3 + 11^3 + 27^3 + 70^3 = 682128

1^3 + 50^3 + 57^3 + 15^3 + 22^3 + 71^3 = 2^3 + 45^3 + 61^3 + 11^3 + 27^3 + 70^3

AND

1 + 50 + 57 + 15 + 22 + 71 = 216 = 2 + 45 + 61 + 11 + 27 + 70

2^2 + 51^2 + 58^2 + 16^2 + 23^2 + 72^2 = 11938

3^2 + 46^2 + 62^2 + 12^2 + 28^2 + 71^2 = 11938

3^2 + 52^2 + 59^2 + 17^2 + 24^2 + 73^2 = 12388

4^2 + 47^2 + 63^2 + 13^2 + 29^2 + 72^2 = 12388

4^2 + 53^2 + 60^2 + 18^2 + 25^2 + 74^2 = 12850

5^2 + 48^2 + 64^2 + 14^2 + 30^2 + 73^2 = 12850

So,

5 + 37 + 66 + 6 + 35 + 67 = 216

1 + 50 + 57 + 15 + 22 + 71 = 2 + 45 + 61 + 11 + 27 + 70 = 216

So 1 + 50 + 57 + 15 + 22 + 71 = 2 + 45 + 61 + 11 + 27 + 70 = 5 + 37 + 66 + 6 + 35 + 67

5^2 + 37^2 + 66^2 + 6^2 + 35^2 + 67^2 = 11500

5^3 + 37^3 + 66^3 + 6^3 + 35^3 + 67^3 = 682128

So,

1^2 + 50^2 + 57^2 + 15^2 + 22^2 + 71^2

= 2^2 + 45^2 + 61^2 + 11^2 + 27^2 + 70^2

= 5^2 + 37^2 + 66^2 + 6^2 + 35^2 + 67^2

1^3 + 50^3 + 57^3 + 15^3 + 22^3 + 71^3

= 2^3 + 45^3 + 61^3 + 11^3 + 27^3 + 70^3

= 5^3 + 37^3 + 66^3 + 6^3 + 35^3 + 67^3

We have 3 sets of integers

(1, 50, 57, 15, 22, 71), (2, 45, 61, 11, 27, 70), (5, 37, 66, 6, 35, 67)

[To Be Continued]

(1) A = (a1, a2, ..., ai) and B = (b1, b2, ..., bj)

(a1) + (a2) + ... + (ai) = (b1) + (b2) + ... + (bj)

(a1)^2 + (a2)^2 + ... + (ai)^2 = (b1)^2 + (b2)^2 + ... + (bj)^2

(a1)^3 + (a2)^3 + ... + (ai)^3 = (b1)^3 + (b2)^3 + ... + (bj)^3

(a1)^4 + (a2)^4 + ... + (ai)^4 = (b1)^4 + (b2)^4 + ... + (bj)^4

(a1)^5 + (a2)^5 + ... + (ai)^5 = (b1)^5 + (b2)^5 + ... + (bj)^5

and,

(a1 + 1)^2 + (a2 + 1)^2 + ... + (ai + 1)^2 = (b1 + 1)^2 + (b2 + 1)^2 + ... + (bj + 1)^2

(a1 + 2)^2 + (a2 + 2)^2 + ... + (ai + 2)^2 = (b1 + 2)^2 + (b2 + 2)^2 + ... + (bj + 2)^2

(a1 + 3)^2 + (a2 + 3)^2 + ... + (ai + 3)^2 = (b1 + 3)^2 + (b2 + 3)^2 + ... + (bj + 3)^2

(a1 + 4)^2 + (a2 + 4)^2 + ... + (ai + 4)^2 = (b1 + 4)^2 + (b2 + 4)^2 + ... + (bj + 4)^2

(2) A = (a1, a2, ..., ai) and B = (b1, b2, ..., bj)

(a1 + p)^2 + (a2 + p)^2 + ... + (ai + p)^2 = (b1 + p)^2 + (b2 + p)^2 + ... + (bj + p)^2

p being the first prime numbers 2, 3, 5, 7

Here are set of equations in which the sums of powers of two different sets of numbers are the same for several different exponents. The simplest example :

1 + 6 + 8 = 15 = 2 + 4 + 9

1^2 + 6^2 + 8^2 = 101 = 2^2 + 4^2 + 9^2

1 + 8 + 10 + 17 = 36 = 2 + 5 + 13 + 16

1^2 + 8^2 + 10^2 + 17^2 = 454 = 2^2 + 5^2 + 13^2 + 16^2

1^3 + 8^3 + 10^3 + 17^3 = 6426 = 2^3 + 5^3 + 13^3 + 16^3

The two sets, here, are: (1, 8, 10, 17) and (2, 5, 13, 16)

Adding 1 to the example above, gives

2 + 9 + 11 + 18 = 40 = 3 + 6 + 14 + 17

2^2 + 9^2 + 11^2 + 18^2 = 530 = 3^2 + 6^2 + 14^2 + 17^2

2^3 + 9^3 + 11^3 + 18^3 = 7900 = 3^3 + 6^3 + 14^3 + 17^3

Adding 2 and 3 to all terms of the equation 1 + 8 + 10 + 17 = 36 = 2 + 5 + 13 + 16

we get,

(1+1)^2 + (8+1)^2 + (10+1)^2 + (17+1)^2 = 530

(2+1)^2 + (5+1)^2 + (13+1)^2 + (16+1)^2 = 530(1+2)^2 + (8+2)^2 + (10+2)^2 + (17+2)^2 = 614

(2+2)^2 + (5+2)^2 + (13+2)^2 + (16+2)^2 = 614

(1+3)^2 + (8+3)^2 + (10+3)^2 + (17+3)^2 = 706

(2+3)^2 + (5+3)^2 + (13+3)^2 + (16+3)^2 = 706

(1+4)^2 + (8+4)^2 + (10+4)^2 + (17+4)^2 = 806

(2+4)^2 + (5+4)^2 + (13+4)^2 + (16+4)^2 = 806

(1+5)^2 + (8+5)^2 + (10+5)^2 + (17+5)^2 = 914

(2+5)^2 + (5+5)^2 + (13+5)^2 + (16+5)^2 = 914

OR

1^2 + 8^2 + 10^2 + 17^2 = 454 = 2^2 + 5^2 + 13^2 + 16^2

2^2 + 9^2 + 11^2 + 18^2 = 530 = 3^2 + 6^2 + 14^2 + 17^2

3^2 + 10^2 + 12^2 + 19^2 = 614 = 4^2 + 7^2 + 15^2 + 18^2

4^2 + 11^2 + 13^2 + 20^2 = 706 = 5^2 + 8^2 + 16^2 + 19^2

5^2 + 12^2 + 14^2 + 21^2 = 806 = 6^2 + 9^2 + 17^2 + 20^2

5^2 + 12^2 + 14^2 + 21^2 = 806 = 6^2 + 9^2 + 17^2 + 20^2

**6^2 + 13^2 + 15^2 + 22^2 = 914 = 7^2 + 10^2 + 18^2 + 21^2**

AND

1 + 50 + 57 + 15 + 22 + 71 = 216 = 2 + 45 + 61 + 11 + 27 + 70

1^2 + 50^2 + 57^2 + 15^2 + 22^2 + 71^2 = 11500

2^2 + 45^2 + 61^2 + 11^2 + 27^2 + 70^2 = 11500

1^3 + 50^3 + 57^3 + 15^3 + 22^3 + 71^3 = 682128

2^3 + 45^3 + 61^3 + 11^3 + 27^3 + 70^3 = 682128

1 + 50 + 57 + 15 + 22 + 71 = 2 + 45 + 61 + 11 + 27 + 70

1^2 + 50^2 + 57^2 + 15^2 + 22^2 + 71^2 = 2^2 + 45^2 + 61^2 + 11^2 + 27^2 + 70^21^3 + 50^3 + 57^3 + 15^3 + 22^3 + 71^3 = 2^3 + 45^3 + 61^3 + 11^3 + 27^3 + 70^3

AND

1 + 50 + 57 + 15 + 22 + 71 = 216 = 2 + 45 + 61 + 11 + 27 + 70

2^2 + 51^2 + 58^2 + 16^2 + 23^2 + 72^2 = 11938

3^2 + 46^2 + 62^2 + 12^2 + 28^2 + 71^2 = 11938

3^2 + 52^2 + 59^2 + 17^2 + 24^2 + 73^2 = 12388

4^2 + 47^2 + 63^2 + 13^2 + 29^2 + 72^2 = 12388

4^2 + 53^2 + 60^2 + 18^2 + 25^2 + 74^2 = 12850

5^2 + 48^2 + 64^2 + 14^2 + 30^2 + 73^2 = 12850

So,

1 + 50 + 57 + 15 + 22 + 71 = 2 + 45 + 61 + 11 + 27 + 70

**1^2 + 50^2 + 57^2 + 15^2 + 22^2 + 71^2 = 2^2 + 45^2 + 61^2 + 11^2 + 27^2 + 70^2**

2^2 + 51^2 + 58^2 + 16^2 + 23^2 + 72^2 = 3^2 + 46^2 + 62^2 + 12^2 + 28^2 + 71^2

3^2 + 52^2 + 59^2 + 17^2 + 24^2 + 73^2 = 4^2 + 47^2 + 63^2 + 13^2 + 29^2 + 72^2

4^2 + 53^2 + 60^2 + 18^2 + 25^2 + 74^2 = 5^2 + 48^2 + 64^2 + 14^2 + 30^2 + 73^2

**5^2 + 54^2 + 61^2 + 19^2 + 26^2 + 75^2 = 6^2 + 49^2 + 65^2 + 15^2 + 31^2 + 74^2**

**6^2 + 55^2 + 62^2 + 20^2 + 27^2 + 76^2 = 7^2 + 50^2 + 66^2 + 16^2 + 32^2 + 75^2**

5 + 37 + 66 + 6 + 35 + 67 = 216

1 + 50 + 57 + 15 + 22 + 71 = 2 + 45 + 61 + 11 + 27 + 70 = 216

So 1 + 50 + 57 + 15 + 22 + 71 = 2 + 45 + 61 + 11 + 27 + 70 = 5 + 37 + 66 + 6 + 35 + 67

5^2 + 37^2 + 66^2 + 6^2 + 35^2 + 67^2 = 11500

5^3 + 37^3 + 66^3 + 6^3 + 35^3 + 67^3 = 682128

So,

1^2 + 50^2 + 57^2 + 15^2 + 22^2 + 71^2

= 2^2 + 45^2 + 61^2 + 11^2 + 27^2 + 70^2

= 5^2 + 37^2 + 66^2 + 6^2 + 35^2 + 67^2

1^3 + 50^3 + 57^3 + 15^3 + 22^3 + 71^3

= 2^3 + 45^3 + 61^3 + 11^3 + 27^3 + 70^3

= 5^3 + 37^3 + 66^3 + 6^3 + 35^3 + 67^3

We have 3 sets of integers

(1, 50, 57, 15, 22, 71), (2, 45, 61, 11, 27, 70), (5, 37, 66, 6, 35, 67)

[To Be Continued]

### Palindrome in multiple bases

Not only am I interested in palindromes which remain palindromes when expressed in a diﬀerent base system, but also in finding a d-digit palindrome base 10 which is palindromic in the greatest number of bases.

## Thursday, March 3, 2011

### Heads or tails?

It all depends on some key variables

http://www.physorg.com/news175267656.html

Everyone knows the flip of a coin is a 50-50 proposition. Only it's not. You can beat the odds. So says a three-person team of Stanford and UC-Santa Cruz researchers. They produced a provocative study that turns conventional wisdom, well, on its head for anyone who has ever settled a minor dispute with a simple coin toss.

http://www.physorg.com/news175267656.html

### March Madness

64 teams play 63 total games

Round:

1 --> 32 games

2 --> 16

3 --> 8

4 --> 4

5 --> 2

6 --> 1

32 + 16 + 8 + 4 + 2 + 1 = 63

There are 2^63 different ways to fill out the bracket.

The current and commonly used formula for determining the RPI of a team at any given time is as follows.

RPI = (WP * 0.25) + (OWP * 0.50) + (OOWP * 0.25)

where WP is Winning Percentage, OWP is Opponents' Winning Percentage and OOWP is Opponents' Opponents' Winning Percentage.

http://en.wikipedia.org/wiki/Ratings_Percentage_Index

Round:

1 --> 32 games

2 --> 16

3 --> 8

4 --> 4

5 --> 2

6 --> 1

32 + 16 + 8 + 4 + 2 + 1 = 63

There are 2^63 different ways to fill out the bracket.

The current and commonly used formula for determining the RPI of a team at any given time is as follows.

RPI = (WP * 0.25) + (OWP * 0.50) + (OOWP * 0.25)

where WP is Winning Percentage, OWP is Opponents' Winning Percentage and OOWP is Opponents' Opponents' Winning Percentage.

http://en.wikipedia.org/wiki/Ratings_Percentage_Index

Is it a good model?

Is March Madness always the same?

March Madness: Statisticians quantify entry biases

http://www.physorg.com/news/2011-03-madness-statisticians-quantify-entry-biases.html

Sports Leagues Not Efficiently Structured, Scientists Say

'Match' Madness: Picking upsets a losing strategy

http://nrich.maths.org/1443

New Algorithm Ranks Sports Teams like Google's PageRankBehavior Changes Linked to March Madness

Is March Madness always the same?

**http://www.physorg.com/news/2011-03-madness.html**

March Madness: Statisticians quantify entry biases

http://www.physorg.com/news/2011-03-madness-statisticians-quantify-entry-biases.html

Odds are, seedings don't matter after Sweet 16, professor says

Computer system consistently makes most accurate NCAA picks

Sports Leagues Not Efficiently Structured, Scientists Say

'Match' Madness: Picking upsets a losing strategy

NCAA Tournament Pool: Leveling The Brackets

Physics Explains Why University Rankings Won't Change

New Algorithm Ranks Sports Teams like Google's PageRank

All bets are off: Office pools lead to unhappiness

## Tuesday, March 1, 2011

### More videos (Gresham College)

This problem is perhaps the perfect introduction to graph theory: How do you connect Mr Angry, Mr Beastly and Mr Cross to gas, water and electricity without the connections crossing one another?

Robin Wilson, Gresham Professor of Geometry, explains the problem and shows how the suitable diagram can prove that a solution is impossible.

4 cubes with 4 colours across their 6 faces, 82,944 possible ways of stacking them - how do we stack them to have all four colours on each side of the stack?

This is the problem of Instant Insanity, to which there are two distinct methods of solution. Robin Wilson, Gresham Professor of Geometry, here explains how to solve it using graph theory.

A short introduction to tilings. Robin Wilson, Gresham Professor of Geometry, runs through some examples of tilings he has met on his travels between Cairo and Colorado.

Dominoes, but with more than two squares... Robin Wilson, Gresham Professor of Geometry, runs through the basic theory of Polominoes and some of its uses, including one made by Arthur C Clarke.

The largest sudoku problem in the world!

Polygonal, Samurai, Samurai Polygonal, Orthogonal, The Dion Cube - Robin Wilson, Gresham Professor of Geometry, runs through the different types of types and varieties of Sudoku problems and the types of challenges that they offer.

The complexity of Magic Squares goes far beyond the Sudoku puzzles we see in our daily newspapers. From 1200BC China, through artists like Duerer, to the master of the 16x16 Magic Square, Banjamin Franklin, Robin Wilson, Gresham Professor of Geometry, gives a quick overview of the possibilities of squares of numbers where the rows and columns add up to the same sum.

What is the average speed if you go to Oxford at 40mph and return at 60mph? - It might not be what you expect! Robin Wilson, Gresham Professor of Geometry, runs through this problem and explains why it the average speed is not 50mph.

Fibonacci is best remembered for his discovery of the 'Fibonacci series'. Robin Wilson, Gresham Professor of Geometry, explains what this series of numbers is and the problem that gave rise to it in his 1202 book 'Liber abaci'.

Two Fibonacci Problems

A fox, a goose and a bag of corn - how do you get them across the river without one of them being eaten...? Robin Wilson, Gresham Professor of Geometry, gives a quick run-through of the origins of the problem and the basis of how to solve it.

The problem Matt Damon's character solved in 'Good Will Hunting' - Homeomorphically Irreducible Trees of degree ten. The problem sounds complex but is actually very easy. Robin Wilson, Gresham Professor of Geometry, explains the problem and shows the simple solutions.

Robin Wilson, Gresham Professor of Geometry, explains the problem and shows how the suitable diagram can prove that a solution is impossible.

4 cubes with 4 colours across their 6 faces, 82,944 possible ways of stacking them - how do we stack them to have all four colours on each side of the stack?

This is the problem of Instant Insanity, to which there are two distinct methods of solution. Robin Wilson, Gresham Professor of Geometry, here explains how to solve it using graph theory.

A short introduction to tilings. Robin Wilson, Gresham Professor of Geometry, runs through some examples of tilings he has met on his travels between Cairo and Colorado.

Dominoes, but with more than two squares... Robin Wilson, Gresham Professor of Geometry, runs through the basic theory of Polominoes and some of its uses, including one made by Arthur C Clarke.

The largest sudoku problem in the world!

Polygonal, Samurai, Samurai Polygonal, Orthogonal, The Dion Cube - Robin Wilson, Gresham Professor of Geometry, runs through the different types of types and varieties of Sudoku problems and the types of challenges that they offer.

The complexity of Magic Squares goes far beyond the Sudoku puzzles we see in our daily newspapers. From 1200BC China, through artists like Duerer, to the master of the 16x16 Magic Square, Banjamin Franklin, Robin Wilson, Gresham Professor of Geometry, gives a quick overview of the possibilities of squares of numbers where the rows and columns add up to the same sum.

What is the average speed if you go to Oxford at 40mph and return at 60mph? - It might not be what you expect! Robin Wilson, Gresham Professor of Geometry, runs through this problem and explains why it the average speed is not 50mph.

Fibonacci is best remembered for his discovery of the 'Fibonacci series'. Robin Wilson, Gresham Professor of Geometry, explains what this series of numbers is and the problem that gave rise to it in his 1202 book 'Liber abaci'.

Two Fibonacci Problems

A fox, a goose and a bag of corn - how do you get them across the river without one of them being eaten...? Robin Wilson, Gresham Professor of Geometry, gives a quick run-through of the origins of the problem and the basis of how to solve it.

The problem Matt Damon's character solved in 'Good Will Hunting' - Homeomorphically Irreducible Trees of degree ten. The problem sounds complex but is actually very easy. Robin Wilson, Gresham Professor of Geometry, explains the problem and shows the simple solutions.

### Lewis Carroll's Problems

Lewis Carroll's Money Problem

How to get the correct change for a shopper's purchase?

A quick introduction to some logical puzzles. Robin Wilson, Gresham Professor of Geometry, runs through some sample problems, from Epimenides to Lewis Carroll.

A quick introduction to some logical puzzles. Robin Wilson, Gresham Professor of Geometry, runs through some sample problems, from Epimenides to Lewis Carroll.

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