Example:
384 = 2^7 * 3 .................................... 385 = 5 * 7 * 11
539 = 7^2 * 11 ................................... 540 = 2^2 * 3^3 * 5
Sunday, March 27, 2011
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
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
(diophantine equations with huge minimal solutions)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?
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
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
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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