Friday, June 17, 2011

Here's Why Our Long-Term Growth Expectations Are Absurd

Investor Jeremy Grantham of GMO illustrates this in his recent apocalyptic commodity analysis with an anecdote about ancient Egypt, which was one of the most successful civilizations in human history.

Grantham asked a group of mathematicians how big Ancient Eqypt would have gotten if its economy had growth 4.5% a year for the 3,000 years the civilization lasted.
The mathematicians were directionally correct: Very big.

But not one of them came even remotely close to the actual number (which is mind-boggling).

I posted on @Quora: Here's Why Our Long-Term Growth Expectations Are Absurd http://qr.ae/7Xjhm #Math #Business

What is the difference between causation and correlation?

http://stats.org/in_depth/faq/causation_correlation.htm

Friday, June 3, 2011

Chess Puzzle

A game begins with 1.e4 and ends in the fifth move with knight takes rook mate.

Tuesday, May 24, 2011

MOBIUS FUNCTION

http://www.gaussianmath.com/misc/mobiusfunction/mobius.html

Introduction to the Möbius Function

Hat 0

Fundamental Theorem of Arithmetic

Uniqueness of Factorization

Mertens Function

Mertens Conjecture

Identities and Mobidromes

Mathematical Wormhole 1 & 2

Time travel is possible in mathematics!

Check out ... http://www.gaussianmath.com/
for a more indepth explanation and other interesting topics in mathematics

The M&M game

It has been said that, "Life is like a box of chocolates—you never know what you're going to get." (Forrest Gump in Forrest Gump, 1994.) In this experiment you can test the "Forrest Gump Chaos Theory" by using M&M's, which are much cheaper than a box of chocolates. What if life is more like a bag of M&M's?

We start with two bags of M&Ms, one with 19 and one with 20.
In her turn, a player has to eat all the M&Ms in one of the bags, and then split the M&M from the other bag between the two, leaving at least one on each bag, and not necessarily evenly. The she gives the two bags to the other player. The player who receives two bags with one M&M each loses as he can no longer move.

Mathematical Platonism

http://www.lehman.edu/deanhum/philosophy/platofootnote/PlatoFootnote.org/Outreach_files/Mathematical%20Platonism.pdf

James Robert Brown’s Philosophy of Mathematics: A Contemporary Introduction to the World of Proofs and Pictures (Routledge, 2008).

There is a difference between general Platonism and the mathematical flavor. For Plato, each apple, say, is but an imperfect example of the absolute (and perfect) Idea of an apple. But as Aristotle quickly realized, Plato has it exactly backwards: we arrive at the general idea of ‘apple’ by mentally abstracting a set of characteristics we think common to all actual apples. It is we who conjure the ‘perfect’ idea from the world, not the world copying the concept.

But now contrast the idea of an apple with the idea of a circle. Here Aristotle’s approach becomes more problematic, as we don’t find any true circles in nature. No natural object has the precise geometric characteristics of a circle, and in a very strong sense we can also say that the circles we draw are but imperfect representations of the perfect idea of a circle. Ah – but whence does such a perfect idea come from?

Consider another way to put the problem. One major difference between science and technology is that science discovers things, while technology is about human inventions. We discover the law of gravity; but we invent airplanes to allow heavier-than-air flight despite the law of gravity. But where do mathematical objects, like circles and numbers, or mathematical theorems like the Pythagorean one, or Fermat’s Last one, come from? Are they inventions of the human mind, or are they discoveries?

Monday, May 23, 2011

Which of the following statements are true?

1. Precisely one of these statements is untrue.

2. Precisely two of these statements are untrue.

3. Precisely three of these statements are untrue.

4. Precisely four of these statements are untrue.

5. Precisely five of these statements are untrue.

6. Precisely six of these statements are untrue.

7. Precisely seven of these statements are untrue.

8. Precisely eight of these statements are untrue.

9. Precisely nine of these statements are untrue.

10. Precisely ten of these statements are untrue.

Visual.ly

visual.ly is launching soon!

Sunday, May 22, 2011

Sum of two dice

2 .................. 1+1 .................................................1/36
3 ...................1+2, 2+1 ..........................................2/36
4 ...................1+3, 2+2, 3+1 ..................................3/36
5 ...................1+4, 2+3, 3+2, 4+1 .........................4/36
6 ...................1+5, 2+4, 3+3, 4+2, 5+1 .................5/36
7 ...................1+6, 2+5, 3+4, 4+3, 5+2, 6+1 ..........6/36
8 ...................2+6, 3+5, 4+4, 5+3, 6+2 .................5/36
9 ...................3+6, 4+5, 5+4, 6+3 ........................4/36
10 ..................4+6, 5+5, 6+4 ................................3/36
11 ..................5+6, 6+5 .........................................2/36
12 ..................6+6 ..................................................1/36

Is it possible to load the dice in such a way that these eleven scores are equally probable?

A clock puzzle

Imagine a clock that does not have any hands or numbers on it, but it has a chimer.

If the time is 1 o'clock, it chimes once. If the time is 2 o'clock, it chimes twice, and so forth. The time gap between any two chimes is 3 seconds.

How many seconds would it take you to know the time, after the first chime is heard, if it is 3 o'clock?

Game show: 5 fortune tellers

On a game show there are 5 fortune tellers (A, B, C, D and E)

A has 81% chance of being correct
B ..... 65% .........................................
C ..... 43% .........................................
D...... 35% .........................................
E ...... 8% ............................................

n times the sum of its digits

Which number is 5 times the sum of its digits?

In general, to find 2-digit and 3-digit numbers that are (2, 3, 4, 5, ...) times the sum of their digits

And any d-digits (d > 3)?

Saturday, May 21, 2011

n circles with radius r ...

... are inscribed in a circle with radius 1

Determine the radius r and the area A of the shaded region between the circles for n = 4 and n = 3

Hint: In an equilateral triangle the medians divide each other in the ratio of 2:1

Sunday, May 15, 2011

Smallest positive integer that includes among its factors perfect squares

1^2 = 1 ..... 2^2 = 4 ..... 3^2 = 9

4^2 = 16 ... 5^2 = 25 ... 6^2 = 36 ... 7^2 = 49 ... 8^2 = 64

10^2 = 100 ..... 11^2 = 121 ..... 12^2 = 144 ..... 13^2 = 169 ..... 14^2 = 196
15^2 = 225 ..... 16^2 = 256 ..... 17^2 = 289 ..... 18^2 = 324 ..... 19^2 = 361
20^2 = 400 ..... 21^2 = 441 ..... 22^2 = 484 ..... 23^2 = 529 ..... 24^2 = 576
25^2 = 625 ..... 26^2 = 676 ..... 27^2 = 729 ..... 28^2 = 784 ..... 29^2 = 841
30^2 = 900 ..... 31^2 = 961

The factors of 2304 are:

1 2 3 4 6 8 9 12 16 18 24 32 36 48 64 72 96 128 144 192 256 288 384 576 768 1152 2304

The prime factors are: 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 3 * 3 or 2^8 * 3^2

2304 is the smallest positive integer that includes among its factors at least 10 perfect squares.

(sqrt(n)^sqrt(n))^sqrt(n) an integer?

http://www.wolframalpha.com/input/?i=%28sqrt%28n%29^sqrt%28n%29%29^Sqrt%28n%29

For how many integral values of n in the range 1 to 1000 (inclusive) is the following statement true?

Prime factoization

Find the smallest positive integer with exactly n di fferent factors

Triangle - Altitude

The figure shows an isosceles triangle ABC with AB = BC

The 80-80-20 Triangle

Is there anything else that could be said about the 80-80-20 triangle?

Problem 358. Isosceles triangle 80-80-20, Circle, Angles, Congruence.
Problem 274: Isosceles Triangle, 80-80-20, Angles

Monday, May 9, 2011

A collection of Math Olympiad problems

(1) Find the smallest natural number which is a multiple of 2009 and whose sum of
(decimal) digits equals 2009.

(2) Find all integer solutions of the equation 3^x - 5^y = z^2

(3) Solve in integers the equation 12^x + y^4 = 2008^z

(4) Determine all pairs of natural numbers (x, n) that satisfy the equation
x^3 + 2x + 1 = 2^n

(5) Determine all pairs (x, y) of integers such that
1 + 2^x + 2^(2x+1) = y^2

(6) Find all primes p such that p^2 – p + 1 is a perfect cube.

(7) Let a, b, c be positive real numbers. Prove the inequality
(a^2 / b) + (b^2 / c) + (c^2 / a) >= a + b + c + 4(a – b)^2 / (a + b + c)
When does equality occur ?

Wednesday, May 4, 2011

A Proposal to End the Practice of Gerrymandering

Using a precise mathematical formula, it is possible to redraw district lines fairly and bring competitiveness back to congressional races

Drawn Together: Can Math Nerds Beat Gerrymandering?

Gerrymandering: Why Your Vote Doesn't Count

Squares of 2 consecutive even numbers using the same digits

206^2 = 42436 and 208^2 = 43264

Find others

Squares and cubes of 2 numbers using the same digits:

102^2 = 10404 and 201^2 = 40401
102^3 = 1,061,208 and 201^3 = 8,120,601.

Digits

202^3 = 8,242,408 = (–8 + 242 – 40 + 8)^3

204 = 3^2 + 7^2 + 5^2 + 11^2 = 2^2 + 8^2 + 6^2 + 10^2,
where 3+7 = 2+8 and after adding 3 to each digit: 6+10 = 5+11.
204^2 = 23^3 + 24^3 + 25^3 = 41,616.

206^3 = 8,741,816 = (8 + 7 + 4 + 181 + 6)^3

208^3 = 8,998,912 = (08 + 99 + 89 + 12)^3

209 = 1^6 + 2^5 + 3^4 + 4^3 + 5^2 + 6^1

305^3 = 28,372,625 = (283 + 7 + 2 + 6 + 2 + 5)^3

494 + 209 = 703 and 494,209 = 703^2

88+209 = 297 and 88,209 = 297^2 and 494+209 = 703 and 494,209 = 703^2 and 297+703 = 1000

2,401 = (2 + 4 + 0 + 1)^4 = 7^4

10,000 = (1+2+3+4)^4 = [(1^5+2^5+3^5+4^5)+(1^7+2^7+3^7+4^7)]/2

Reversible numbers and their squares:

102^2 = 10404 and 201^2 = 40401.

103^2 = 10609 and 301^2 = 90601.

Square numbers using the same digits:

108^2 = 11664, 129^2 = 16641 and 204^2 = 41616

148^2 = 21904, 203^2 = 41209 and 302^2 = 91204.

102^2 = 10404 ... 120^2 = 14400 ... 201^2 = 40401 ... 210^2 = 44100.

148^2 = 21904 ... 203^2 = 41209 ... 302^2 = 91204.

130^2 = 16900 ... 140^2 = 19600 ... 310^2 = 96100,
103^2 = 10609 ... 247^2 = 61009 ... 301^2 = 90601.

207^2 = 42849, 222^2 = 49284 and 288^2 = 82944

128^2 = 16,384, 178^2 = 31,684, 191^2 = 36,481, 196^2 = 38,416 and 209^2 = 43,681.

Monday, May 2, 2011

Math Education

TWO TYPES OF MENTAL ARITHMETIC
AND THE EMPTY NUMBERLINE

tan(4*pi/3) and arctan(4*pi/3)

tan(4*pi/3)

tan(4*pi/3) is clearly an irrational number. How to that arctan(4*pi/3), in radians, is irrational?

Wednesday, April 27, 2011

Rectangle and Circle

Suppose we have a circle inscribed within a square. Now suppose a rectangle extends from one corner of the square to a point on the circle (see diagram).

If this rectangle is 6 inches by 12 inches, what is the radius of the circle in inches?

Solution

Rectangle, Circles inscribed

In a 5 by 12 rectangle, one of the diagonals is drawn and circles are inscribed in both right triangles thus formed. Find the distance between the centers of the two circles.

UPDATE!

Triangle Formulas : Right Triangle
C = A + B = Pi/2 radians = 90 degrees
c^2 = a^2 + b^2
P = a + b + c
s = (a+b+c)/2
K = ab/2

Particular case: Circle Inscribed in Triangle 3, 4, 5

Circle Inscribed in a Right Triangle

General case: Circle Inscribed in Triangle a, b, c
We can write,
r = ab/(a + b + c) with c = sqrt(a^2 + b^2)
d = sqrt((a - 2r)^2+(b - 2r)^2)

OR

r = (a + b - c)/2 where c = Sqrt(a^2 + b^2)
Then (a - 2r)^2 + (b - 2r)^2 = (c - a)^2 + (c - b)^2 etc

Math articles #3

Some horrible functions

24

Using only the operations add, subtract, multiply or divide can you always make 24 with two 3s and one other pair of digits?

For example using 3355, you can make 24 by saying 5x5 –(3/3) and with the digits 3399 it is (3+9) + (3+9)

Using only addition, subtraction, multiplication and/or division make 24 using the four digits 3377 and 3388.

3333?
3344?

Find a 2-digit number such that

Find a 2-digit number such that its square and its fifth power contain together all the digits from 1 to 9, each once and only once.

Numbers that are equal to the sum of the squares of their two "halves"

1233 = 12^2 + 33^2

8833 = 88^2 + 33^2

10100 = 10^2 + 100^2

5882353 = 588^2 + 2353^2

Could you find other examples?

Tuesday, April 26, 2011

Transposable integer

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

Given a positive integer k another positive integer x is said to be k-transposable if, when its leftmost digit is moved to the unit's place, the resulting integer is k*x

For example, the integer 142857 is 3-transposable since

428571 = 3 * 142857

σ(p^4)

The only prime p such that σ(p^4) is a square is 3.

σ(p^4) = 1 + p + p^2 + p^3 + p^4

when p = 3, 1 + 3 + 3^2 + 3^3 + 3^4 = 11^2

Multiples of their reversals

How many integers less than 10000 are multiples of their reversals?

For example, 8712 = 4 * 2178

Monday, April 25, 2011

3 Circles

The figure below shows three circles, a square, and some tangents:

The two lower circles have the same size. Do all three have the same size?

Matching even and odd numbers

Match each of the even numbers between 2 and 30 with a different odd number between 1 and 29 so that the fifteen resulting sums are pairwise relatively prime.

Cycloid

A cycloid is the curve generated by a point on a wheel as the wheel rolls. This diagram shows half of a loop of the standard cycloid, whose parametric form is
(t - sin t, 1 - cos t):

Which is larger, the blue area or the orange area?

2 semicircles, 1 isosceles triangle

Consider two semicircles S and T that emanate from the same point with diameters along a common line, with S being the larger.

Draw an isosceles triangle whose base is the part of diameter of S that is outside T and whose apex is on S. Draw a circle inside S that is tangent to S, T, and the triangle.

Prove that the center of this circle is directly above the point common to the triangle and T

88

Can you get 88 using only the integers 1, 2, and 3, using each of them only once, and using the standard arithmetic operations, including factorials and repeating decimals?

RoundUp

Alice and Bob play a game. A positive integer starts the game and the players take turns changing the current value and passing the new number back to their opponent.

On each move, a player may subtract 1 from the integer, or halve it, rounding up if necessary. The person who first reaches 0 is the winner.

Alice goes first: she makes her choice of move on the starting value.

For example, starting at 15 a legal game (if not particularly well played) could be:

Alice .......................................15 → 8
Bob ......................................... 8 → 7
Alice ....................................... 7 → 4
Bob ......................................... 4 → 2
Alice ....................................... 2 → 1
Bob ......................................... 1 → 0 Bob wins.

For which values of n is there a winning strategy for Alice?

RE: 1 message

I can't view that message.

Squares of prime numbers

X is a square of a prime number, and X > 41

Prove that (X + 41)(X - 41) is evenly divisible by 240

Sunday, April 24, 2011

Limit sin x°/x°, x -> 0

Surprisingly, many get it wrong, when asked :

What is the

sin x°/x°, x -> 0

Dividing a cube

For what smallest k is it possible to divide a cube onto k non-overlapping tetrahedrons?

Lucky numbers

A bus ticket is considered to be lucky if the sum of the first three digits equals to the sum of the last three (6 digits in Russian buses).

Prove that the sum of all the lucky numbers is divisible by 13.

***** UPDATE! *****

Mr. Beefy,
I've just found these 2 sites (it saves me the trouble of typing)

The smallest square number such that...

Find the smallest exact square with last digit not 0, such that after deleting its last two digits we shall obtain another exact square

Common divisor of (a+b) and (a^2 + b^2)

Natural numbers a and b are relatively prime.

Prove that the greatest common divisor of (a+b) and (a^2 + b^2) is either 1 or 2

((x-y)^5+(y-z)^5+(z-x)^5) is divisible by 5(x-y)(y-z)(z-x)

Given x,y,z, three different integers.

Prove that ((x-y)^5 + (y-z)^5 + (z-x)^5) is divisible by 5(x-y)(y-z)(z-x)

Maximal area of a triangle

What maximal area can have a triangle if its sides a,b,c satisfy inequality

0<= a<=1<= b<=2<= c<=3?

(2n + 1)^n >= (2n)^n + (2n - 1)^n

Prove that for every natural n the following inequality is held:

(2n + 1)^n >= (2n)^n + (2n - 1)^n

a^3 + b^3 + c^ 3 + 3abc > ab(a+b) + bc(b+c) + ac(a+c)

Prove that for all the positive numbers a,b,c the following inequality is valid:

a^3 + b^3 + c^ 3 + 3abc > ab(a+b) + bc(b+c) + ac(a+c)

Median and Average Price

A real estate website reported that the median price of single family homes sold
in the past 9 months in the local area was \$136,900 and the average price was \$161,447.

How do you think these values are computed?

Which do you think is more useful to someone considering the purchase of a home, the median or the average

Saturday, April 23, 2011

Eight queens puzzle

The original problem:
The eight queens puzzle is the problem of placing eight chess queens on an 8×8 chessboard so that no two queens attack each other.
http://en.wikipedia.org/wiki/Eight_queens

In this version, we need to place as many queens as possible on an NxN board, such that each queen will threaten at most ONE other queen. We ask to prove an upper bound and give a solution matching it for the standard 8x8 board as well as for a 30x30 board.

(1 + 2cos(20 °))^n

Find a natural number n for which the expression round (1 + 2cos(20 °))^n is divisible by 1,000,000,000.

***** UPDATE! *****

The function:

Mr. Beefy,
If n = 35

If n = 68

I need to review my work on this. I wanted to express this as a polynomial of cos(x).

I started with the basic identities cos(2x), sin(2x) and cos(3x)

Dominoes

Prove: if we remove two opposite corners from the chessboard, the board cannot be covered by dominoes. (Each dominoe covers two neighboring cells of the chessboard.)

Look for an “Ah-ha” proof: clear, convincing, no cases to distinguish.

Edos test

Let S be a set of n + 1 integers from {1, 2, . . . , 2n}.
Prove that two of them are relatively prime.
First show that this can be avoided if S has only n numbers

Balancing numbers

Suppose we have 13 real numbers with the following property: if we remove any one of the numbers, the remaining 12 can be split into two sets of 6 numbers each with equal sum.

Prove: all the 13 numbers are equal.

(Hint: ﬁrst assume all the numbers are integers.)

RUBIK'S PUZZLES

Rubik's Clock
Objective: Get all clock's at both sides at 12 o'clock. By pushing or pussing one of the four buttons different clocks will move at the same time.

Rubik's Dice
Object: Re-arrange the metal plates inside the cube in such a way that the dice has only white spots. If red is shown anywhere on the dice even through the small controll holes, the puzzle is not complete. The number of possible combinations is 7! x 4^7=82,575,360. There is only ony one correct solution

Friday, April 22, 2011

Barry Nalebuff on Poker and Game Theory

Question: What is the relationship between game theory and poker?

Barry Nalebuff: There’s a great example of game theory and poker that is from an economist colleague of mine named Randy Heeb. He teaches strategy and sometimes teaches with me at Yale.

It was I think 2002 when the World Championship, he was playing a fellow named Thomas Preston, who was probably better known as “Amarillo Slim” and they were playing the World Championship Heads Up, which is a form of Texas HoldEm.

It was €40,000 pot, which this day may seem small, but he was assistant professor, so it’s real money. And it turns out the top two people were Slim and my friend Heeb. Heeb have a pair of Knights, and based on the betting and the initial hands, he was convinced that Slim either had a King or an Ace on his hand, but not both.

Now, what happens next is three cards, or the flop, are presented, and then there is another round of betting. And Heeb had decided that if either a King or an Ace came out in the flop, he was going to fold, because basically the chances of the pair of Kings or Aces, which is too high. And clearly if Slim had a pair of King or Aces, Heeb stay in, he was going to leave with that kind of a hand.

The cards came out and there’s a 2 of Clubs and 3 of Diamonds and a King of hearts. And at that point, Heeb sort of realized, damn it, I’ve got big pot, I’m going to lose it, and his eyes just gave it away. He realized that he had not done a good job hiding his emotions, and at the same time, he also realized that Slim had looked into his eyes seeing that dejection and went all in.

At that point, Heeb said, “Ah, Slim wasn’t betting based on his cards. He was betting based on the fact that I was giving in, and therefore it means he had the Ace not the King, so he doesn’t have the pair and so he went all in and won.”

And it was this great analysis of, okay, what is he thinking, because if you truly have the pair then what he is happy about is the cards, not about my rejection and that wasn’t what Slim was betting on.

But, typically, poker players spend a lot more time watching the other players than they do watching their own cards and also that’s why a lot of people wear sunglasses.

How to Get Ahead Using Career Game Theory

Stephen Miles: When you interview people and you ask them about their careers retrospectively what most people will say is stuff just happened to them. “I was here.” “I was lucky.” They’ll attribute it to all sorts of things

If you think about your career and think about it broadly in terms of what am I learning, what is this position doing for me, how am I going to get to the next one, what is the most important next move for me and you think two or three moves ahead you can start to put yourself in advantageous positions to accelerate your career disproportionately. As opposed to sort of creeping along the floor you can start to sprint.

Before you set out on your own personal game theory you have to assess your own playing field and who is on the playing field:

Does your boss want to be known for creating talent and being a net contributor of talent inside your company or organization or does your boss want to be known for really delivering outstanding results because although subtle it really matters for you as an employee because if they want to be a net contributor of talent they’re willing to rotate people through positions, coach mentor them and then if you will, set them free in other parts of the organization. If they’re solely focused on delivering results then what they’re going to do is over hire people for the job and keep them in the job for as long as they possibly can because that is how you’re going to deliver results. You’re not going to take any risk on people. So you as a person playing on that field needs to think about what are your boss’s motives and what are they known for.

There is only one job, so you’re competing for that job with your peers and you have to assess what are their strengths, what are they doing that is disproportionate to what you’re doing and how do you position yourself to be the best candidate for the next job, so you have to assess horizontally if you will, the peer playing field.

If they choose you to go to the next job, do they have somebody to backfill? Have you created a successor for yourself because if you haven’t, if you’ve made yourself indispensible, guess what, you’re probably going to be indispensible and, therefore, not moved. So you have to think about below are you creating the conditions to make it easy for you to move and allow your boss to make that move with you?

Drilling a square hole!

Here we are using a shape of constant width to drill a square hole!

Hart's straight line

This is a mechanism which will draw a perfect staight line. Can you work out why it does this?

The first planar linkage which draws an exact straight line was invented by Charles Nicolas Peaucellier (1832-1913) in 1864.

The two-tip tetrahedron

This tetrahedron has only two stable faces.

Rolling discs

Take two identical discs and slot them together along diameters so that the discs are perpendicular. There is one separation of centres for which the compound shape has a centre of mass which remains a constant height above the surface of a table. The resulting shape rolls along in a most intriguing way.

Historically, the first straight line linkage was described by Sarrus in 1853. It differs in that its parts move in three dimensions. It is applied widely in jacks, elevating platforms and similar devices.

The uni-stable polyhedron

Prof. John Conway proved that there are uni-stable polyhedra. That is to say, polyhedra which are stable on only one face.

The super egg

This "super egg" stands on one end - but looks as if it should fall over!

Solids of constant width

The width of a smooth shape is the distance between parallel tangents. When a shape is not smooth we must talk about parallel support lines instead. If we have a circle then the width is constant. If the width is constant, then is the shape a circle? These solids have constant width.

Evil wizard, 2 identical vases, beads

You have been captured by an evil wizard. However the wizard also likes maths, and sets you a challenge. You have two identical vases, and 100 white beads and 100 black beads. You must arrange the beads however you like between the two vases. The only condition is that no vase can be empty. And all the beads must be in the vases.

The wizard will then get his assistant to choose a single bead from one of the vases. The assistant will pick purely at random, and will not peek! If the assistant picks a black bead, you will go free. If the assistant picks a white bead... well, let's not go there. Obviously you would like the assistant to pick a black bead.

The question is this. How do you arrange the beads so as to give yourself the best chance of freedom?

There are no tricks. Just mathematics.

The answer will be posted next week

Thursday, April 21, 2011

Pick any four positive integers

For example, 5, 14, 17, and 23 give a collection of four positive integers. Now, form the differences of pairs of these numbers by subtracting the smaller number from the larger in each pair (although this won’t really matter). In our case, this process would give us the following:

(14 – 5) = 9 ...................... (17 – 14) = 3

(17 – 5) = 12 .................... (23 – 14) = 9

(23 – 5) = 18 .................... and (23 – 17) = 6

Now, form the product of all these differences. This gives us 9 x 12 x 18 x 3 x 9 x 6 = 314,928, in our case. Let’s call the result the Prod-Dif of the four integers. So, the Prod-Dif of the collection {5, 14, 17, 23} is 314,928. We can form the Prod-Dif of any collection of four positive integers. Then we have a very interesting problem to ponder:

What is the largest integer that divides evenly into all Prod-Difs? That is, what is the greatest common factor of the collection of all Prod-Difs? And of course, WHY?

Wednesday, April 20, 2011

Bejeweled Blitz

Question : Can the board be set such that no legal move is possible?
=======

Bejeweled Blitz is a game played on an 8x8 space with 6 different color gems.
See the picture below.

The object of the game is to move the gems so that combinations are formed, hence gain points.

1) The board is set such that the same color gem does not occupy 3 consecutive spaces in any column or row.

2) A legal move consists of exchanging 2 neighboring gems (horizontally or vertically NOT diagonally) such that combinations of 3 or more are formed.

In the picture above, the highlighted yellow gem can exchange positions with the purple above it to form a combination of 3 horizontal yellows. There are other possible moves e.g. moving white to the bottom right position to form a combination of 3 vertical whites

What number can be expressed as sum of 6 squares

What is the smallest number that can be expressed as a sum of six different squares but cannot be expressed as the sum of five different square numbers?

UPDATE!

Take,

1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2 = 91

Could you express 91 as a sum of five different square numbers?

The six sixes problem

Please complete the list (Only the digit 6 is allowed):

1 =
2 =
3 =
4 =
5 =
...........
...........

240 =
241 = ((.6+((6+6)*(6+6)))/.6)
242 = ((6*(6+(6*6)))-(6/.6))
243 = (6+((6*(.6*66))-.6))
244 = (.6...*(6+(6*(66-6))))
245 = ((((6)!+((6)!+66))/6)-6)
246 = (66+(6*((6*6)-6)))
247 = (66+((6+((6)!/.6...))/6))
248 = (6*(6+(6*(6-(.6.../6)))))
249 = (.6+(6*(6+((6*6)-.6))))
250 = (((6*(6*6))-66)/.6)
251 = ((6*(6+(6*6)))-(6/6))
252 = (66+(66+((6)!/6)))
253 = ((6/6)+(6*(6+(6*6))))
254 = ((.6...*((6*66)-6))-6)
255 = ((((6*6)+66)/.6)/.6...)
256 = (6*(6*(6-(6/(.6-6)))))
257 = (6+(((6)!+((6)!+66))/6))
258 = ((6)!-(66+(6*66)))
259 = ((((6*6)+((6)!/6))-.6)/.6)
260 = ((66+(((6)!/.6)/6))-6)

The five fives problem

Please complete the list: (only the digit 5 is allowed)

1 =
2 =
3 = ((55 / 5) - 5) * .5
4 = (5 / (5 / 5)) - (5 / 5)
5 = 5 / (55 / 55)
6 = (5 - (5 / 5) - (5 / 5)) !
7 = 5 + 5/5 + 5/5
8 = (5 + (55 / 5)) * .5
9 =
10 = ((5 !) / (5 + (5 / 5))) * .5
11 =
12 =
13 = ((5 !) / 5) - (55 / 5)
14 = 5 + 5 + 5 - (5 / 5)
15 = .5 * (5 + 5/5) * 5
16 = 5 + 5 + 5 + (5 / 5)
17 =
18 =
19 = ((5 !) / 5) - (5 / .5) + 5
20 =
21 =
22 =
23 = ((5 - (5 / 5)) !) - (5 / 5)
24 = 5 * 5 - (5/5)^5
25 = (5 / .5) + (5 / .5) + 5
26 = 5 * 5 + (5/5)^5
27 =
28 =
29 = ((5 !) / 5) + (5 / .5) - 5
30 = ((55/5) - 5) * 5
31 =
32 = ((5/5) + (5/5))^5
33 =
34 = ((5 - (5 / 5)) !) + (5 / .5)
35 =
36 =
37 = (((5 + 5) / 5)^5) + 5
38 =
39 = ((5 !) / 5) + (5 / .5) + 5
40 =
41 =
42 =
43 = ((5 !)/5) + ((5 !)/5) - 5
44 = 55 - (55 / 5)
45 = 5 * (5 + 5 - (5 / 5))
46 =
47 =
48 =
49 = (.5 * (5 !)) - (55 / 5)
50 =
51 =
52 =
53 = ((5!)/5) + ((5!)/5) + 5
54 =
55 =
56 =
57 =
58 =
59 =
60 = (5 !) / ((5 / 5) + (5 / 5))
61 =
62 =
63 =
64 =
65 =
66 = (55 / 5) + 55
67 =
68 =
69 =
70 =
71 = (.5 * (5 !)) + (55 / 5)
72 =
73 =
74 =
75 =
76 =
77 =
78 =
79 =
80 = 5 * (5 + (55 / 5))
81 =
82 =
83 =
84 =
85 =
86 =
87 =
88 =
89 =
90 =
91 =
92 =
93 =
94 =
95 =
96 = (((5 !) / .5) + ((5 !) / .5)) / 5
97 =
98 =
99 =
100 = 5 * (5 - (5 / 5)) * 5
101 =
102 =
103 =
104 =
105 =
106 = (555 / 5) - 5
107 =
108 =
109 =
110 =
111 =
112 =
113 =
114 =
115 =
116 = (555 / 5) + 5
117 =
118 =
119 =
120 = (5 + (5 / 5) - (5 / 5)) !
121 =
122 =
123 =
124 =
125 =
126 =
127 =
128 =
129 =
130 = ((5^5) / (5 * 5)) + 5
131 =
132 =
133 =
134 =
135 =
136 =
137 =
138 =

139 = ((((5 + (5/5)))!/5) - 5)
140 = (.5*(5 + (5*55)))
141 = ((5)! + ((5 + (5 +.5))/.5))
142 = ((5)! + ((55/.5)/5))
143 = ((((5 + (5/5)))! - 5)/5)
144 = ((((55/5) - 5))!/5)
145 = ((5*(5 + (5*5))) - 5)
146 = ((5)! + ((5/5) + (5*5)))
147 = ((5)! + ((.5*55) - .5))
148 = ((5)! + (.5 + (.5*55)))
149 = (5 + (((5 + (5/5)))!/5))

Using only 3,4,5 straight lines to cut a pizza into k pieces

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Explanation: A(n) = n(n + 1)/2 + 1

1 Thru 2000 Factorial

1! = 1
2! = 2
3! = 6
4! = 24
5! = 120
6! = 720
7! = 5040
8! = 40320
9! = 362880
10! = 3628800
11! = 39916800
12! = 479001600
13! = 6227020800
14! = 87178291200
15! = 1307674368000
16! = 20922789888000
17! = 355687428096000
18! = 6402373705728000
19! = 121645100408832000
20! = 2432902008176640000
21! = 51090942171709440000
22! = 1124000727777607680000
23! = 25852016738884976640000
24! = 620448401733239439360000
25! = 15511210043330985984000000
26! = 403291461126605635584000000
27! = 10888869450418352160768000000
28! = 304888344611713860501504000000
29! = 8841761993739701954543616000000

Tuesday, April 19, 2011

Sqrt(5), Sqrt(7),..., Sqrt(k)

Given a unit square and a straight edge, what's the minimum number of lines you'd need to construct to produce a line that's

Sqrt(5), Sqrt(7), ..., Sqrt(k)

Beyond Erdos conjecture

Brocard's Problem:

n! + 1 = m^2

Erdos conjectured that (5,4),(11,5),(71,7) are the only 3 pairs.

Now, I ask : how many solutions to

n! = x^2 + y^2

are there?

0!, 1!, 2! and 6! can be written as sum of two perfect squares. Any others?

The smallest and largest square numbers

The smallest and largest square numbers containing the digits 1 to 9 are :

11826^2 = 139854276 ..... 30384^2 = 923187456

The smallest and largest square numbers containing the digits 0 to 9 are:

32043^2 = 1026753849 ..... 99066^2 = 9814072356

Aurifeuillian Factorizations

n^4 + 4 is composite (i.e., not prime) for all n > 1

Well, n^4 + 4 = (n^2 - 2n + 2) (n^2 + 2n + 2) is a proper factorization, since the smaller of the two factors is greater than 1 when n > 1.

This type of factorization is often called Aurifeuillian (sometimes also spelled Aurifeuillean) in honor of the French mathematician [Léon François] Antoine Aurifeuille.

For example, the following Aurifeuillian (preliminary) factorization often pops up when x is a power of 2:

4 x^4 + 1 = (2x^2 - 2x + 1)( 2x^2 + 2x + 1)

If x = 2y^3, then:

(2 x^2 + 2x + 1) = ( 2 y^2 - 2y + 1)(4 y^4 + 4 y^3 + 2 y^2 + 2y + 1)
(2 x^2 - 2x + 1) = ( 2 y^2 + 2y + 1)(4 y^4 - 4 y^3 + 2 y^2 - 2y + 1)

In either case, the first factor has a similar factorization when y = 2z^3 etc.

Thus, we have two preliminary factors of 2^n+1 when n is congruent to 2 modulo 4, we've 6 of them when n is congruent to 6 modulo 12, and so forth...

4 a^4 + b^4 = (2 a^2 - 2ab + b2 )(2 a^2 + 2ab + b^2 )
27 a^6 + b^6 = (3 a^2 - 3ab + b2 )(3 a^2 + b^2)(3 a^2 + 3ab + b^2)

Monday, April 18, 2011

TV Show NUM3ERS (Puzzles)

Rearrange the key caps of the 1 through 9 on a numeric keypad so that no cap is on its correct key, in such a manner that each of the three rows forms a 3-digit perfect square.

Math in baseball: Ruth-Aaron pair

when Hank Aaron's 715th home run beat Babe Ruth's record of 714,
Carl Pomerance noted that

714 = 2*3*7*17 and 715 = 5*11*13

contained the first seven primes and the sums were each 29:
2 + 3 + 7 + 17 = 5 + 11 + 13 = 29

He called n a Ruth-Aaron number if the sum of its prime divisors, counting multiplicity,
was the same as that for n + 1

Pomerance named pairs like 714-75 Ruth-Aaron pairs, and calculated all the pairs below 20,000.

He also conjectured that this kind of pairs occurred infinitely often, but have no idea of how to prove this when he published this in the JRM.

The number of them that are less than x has been shown by Pomerance to be

O (x(ln lnx)^4 / (ln x)^2)

One week after the publication Paul Erdos got the proof of the infinitude of Ruth Aaron pairs

See here

Now an ugly issue: what happens if in the prime decomposition of the numbers involved in the Ruth Aaron pairs, one or several of these prime are powered? Will you take all the prime factors involved (with repetitions) or will you take only the distinct primes involved (without repetition)?

As a matter of fact you can do it in one way or another, with the result that you will generate two kind of sequences: Ruth-Aaron pairs, prime factors

(a) with repetition http://oeis.org/A039752

(b) without repetition http://oeis.org/A006145

Ruth-Aaron triplets do exist.
That is to say, three consecutive numbers, n, n+1 and n+2 such that the sum of the prime factors of each number adds up to the same quantity a) without repetition or b) with repetition.

The first example, prime factors with repetition, is the triplet 417162, 417163 and 417164:

417162 = 2x3x251x277
417163 = 17x53x463
417164 = 2x2x11x19x499

2+3+251+277 = 17+53+463 = 2+2+11+19+499 = 533

The first example, prime factors without repetition, is the triplet 89460294, 89460295, 89460296:

89460294 = 2x3x7x11x23x8419
89460295 = 5x4201x4259
89460296 = 2x2x2x31x43x8389

2+3+7+11+23+8419 = 5+4201+4259 = 2+31+43+8389 = 8465

Question:

Can you find three more example of each kind?

Ruth-Aaron pairs revisited

Geometry gems

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Sunday, April 17, 2011

Trigonometry

What is

sin 1° + sin 2° + sin 3° + ... + sin 90°

And what is the product

sin 1° * sin 2° * sin 3° * ... * sin 90°

Multiplicatively Perfect numbers

The number of positive multiplicatively perfect numbers less than 100

2519 Mod n (n = 2, ... 10).

2519 Mod 2 = 1 .............. 2519 = 1259 * 2 + 1
2519 Mod 3 = 2 .............. 2519 = 839 * 3 + 2
2519 Mod 4 = 3 .............. 2519 = 629 * 4 + 3
2519 Mod 5 = 4 .............. 2519 = 503 * 5 + 4
2519 Mod 6 = 5 .............. 2519 = 419 * 6 + 5
2519 Mod 7 = 6 .............. 2519 = 359 * 7 + 6
2519 Mod 8 = 7 .............. 2519 = 314 * 8 + 7
2519 Mod 9 = 8 .............. 2519 = 279 * 9 + 8
2519 Mod 10 = 9 ............. 2519 = 251 * 10 + 9

Logic Riddles (videos)

Sarah's Certain Death Riddle
From the Labyrinth, this riddle seems to stump everyone!

Expressions with e, i and π

We all know that ... e^(i*π) + 1 = 0

I found out that ....

e^(i*π/3) + e^(2i*π/3) + e^(3i*π/3) + e^(4i*π/3) + e^(5i*π/3) + e^(6i*π/3) = 0

And

e^(i*π/3) + 2e^(2i*π/3) + 2e^(3i*π/3) + 2e^(4i*π/3) + e^(5i*π/3) + 9e^(6i*π/3) = 6

e^(2i*π/3) + e^(4i*π/3) + e^(6i*π/3) = 0

6e^(6i*π/3) = 6

6*[cos(2π) + i*sin(2π)] = 6

6/π^2

It's also the probability that a number picked at random from the set of integers will have no repeated prime divisors

e^(2i*π) = e^(4*i*π) = e^(6*i*π) = 1

e^(3i*π) = -1 .... e^(5i*π) = -1 .... e^(7i*π) = -1

e^(2!*i*π) = e^(3!*i*π) = e^(4!*i*π) = e^(5!*i*π) = 1

Tossing 2 biased coins

2 coins : coin A and coin B.

Coin A: "2/3 - 1/3" that is to say, it has a 2/3 chance of landing Heads and a 1/3 chance of landing Tails.
Coin B: "1/4 - 3/4"

The first coin tossed is Coin A.

What is the probability that you see two heads in a row before you see two tails in a row?

Saturday, April 16, 2011

What is 0^0 ?

http://mathforum.org/dr.math/faq/faq.0.to.0.power.html

Read also: From the Dr. Math archives:
Why are operations of zero so strange

Perfect number

A perfect number is ... http://en.wikipedia.org/wiki/Perfect_number

6, 28, 496, 8128, 33550336, 8589869056, 137438691328, 2305843008139952128, 2658455991569831744654692615953842176, 191561942608236107294793378084303638130997321548169216

Question :

Show that any even perfect number greater than 28 can be represented as the sum of at least two perfect numbers.

Palindromic number on the Odometer

Too bad, the number on the picture is not a palindromic number.

Suppose the odometer of your car shows a palindromic number, say, 15, 951. And exactly two hours later, the odometer shows a new palindromic number.

What was the average speed of the car in those two hours?

Prime palindromes with this property

An odd-odd number is a positive integer all of whose digits are odd.

How many integers < 10,000 are prime palindromes but not odd-odd?

Pascal's triangle

(1) Find all instances of three consecutive terms in a row of Pascal's triangle
in the ratio 1 : 2 : 3

(2) If we replace even integers by 0 and odd integers by 1 in the Pascal's triangle,
we get a modulo 2 Pascal triangle

Will 1101 or 1011 occur as a consecutive segment in any row of this modulo 2 Pascal triangle?

Palindromic Pentagonals

http://www.worldofnumbers.com/penta.htm

Numbers that are not the sum of 3 pentagonal numbers http://oeis.org/A003679