## 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

***********************************

***********************************

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

*********************************************

*******************************************

## 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

### Palindromic number records

A 53-digit Palindromic Square Number:

122063831551139898460740721 ^ 2 = 14899578972945056149893218681239894165054927987599841

Other palindromic number records can be found at

### Maths in the news #4

O'Dorney Wins Intel Science Talent Search

### Pentagonal numbers

Triangular number

Prove that every pentagonal number greater than one can be written as the sum of three triangular numbers, two of which are equal.

Triangular numbers T_n are positive integers of the form n(n + 1)/2

Pentagonal numbers P_n are positive integers of the form n(3n - 1)/2

@matthen2 posted a picture proof http://twitpic.com/4li960

## Friday, April 15, 2011

### Palindromic number 698,896

The number 698,896 (a palindrome) is the square of 836. Is this the only square palindrome containing an even number of digits?

Well, no. Here are other Palindromic Squares

228138929476341405^2 = 52047371142611077177011624117374025
404099764753665981^2 = 163296619873968186681869378916692361
2282211769458230805^2 = 5208490560653668833388663560650948025

N.B. More research is needed. There are others that escape detection!

### Triangular numbers and Oblong numbers

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

1 + 2 + 3 + ... + n = n(n + 1)/2

And adding the successive odd numbers, we get

1 + 3 + 5 + ... + (2n - 1) = n^2

And if we sum the series of even numbers

2 + 4 + ... + 2n = n(n + 1)

8 * n(n + 1)/2 + 1 = 4n^2 + 4n + 1 = (2n + 1)^2
( 8 times any triangular number + 1 makes a square )

The difference between the squares of any two consecutive triangular numbers is always
a cube:
http://mathworld.wolfram.com/TriangularNumber.html

The oblong number is double of a triangular number.

The successive oblong numbers are:
2*3 = 6, 3*4 = 12, 4*5 = 20, ... n(n + 1), ...

When is the Product of Two Oblong Numbers Another Oblong?

Question :

Show that Gauss's discovery that every number is the sum of three or fewer triangular numbers implies that every number of the form 8k + 3 can be expressed as the sum of three odd squares

### Number Line Tightrope

How slinkies get entangled, and how to untangle them?

To get a better feel for it, get hold of a beaded necklace (like the ones they fling around at Mardi Gras). Lay it out flat and straight. Twist the two strands together at one point, then let go. They'll stay together. If you try to pull them apart, likely nothing will happen; you have to untwist them in exactly the same way, at the same point. Now, do the same thing with a slinky. The reason that undoing it again is harder, is that where they intertwine wanders up and down the slinky as the spirals shift against each other. However, if you pull apart what you can, you'll find that there are a few specific points (or, in the case you're describing, one specific point) where they're actually locked together. Treat each end of the slinky, from the last knot onwards, like a giant bead. Twist them around each other in the right direction, and they'll come right apart.

Any other ideas?

### AP Stats hypothesis test mistakes

What it looks like when everything is done wrong.

### The p-value

A conversation between a clinical researcher and a statistician about the meaning of a p-value (this is loosley based on the beginning of an article, "What your statistician never told you about p-values", by Blume and Peipert).

Reference:

Blume, J. and J. F. Peipert (2003). "What your statistician never told you about P-values."
J Am Assoc Gynecol Laparosc 10(4): 439-444.

Power of the test, p-values, publication bias and statistical evidence
A discussion of statistical evidence and why you might not get results significant enough to reject your null hypothesis even if your alternative hypothesis is correct.

This video is related to the article "Statistics notes: Absence of evidence is not evidence of absence" by Douglas G. Altman and J. Martin Bland. Video created by Matt Asher. For more statistical insights please visit statisticsblog.com

### How not to collaborate with a biostatistician

How not to collaborate with a biostatistician. This is what happens when two people are speaking different research languages! My current workplace is nothing like this, but I think most biostatisticians have had some kind of similar experiences like this in the past!

### James Galea (Card Trick) Comedy Festival Gala 2009

James Galea at The 2009 Melbourne International Comedy Festival Gala for Oxfam Australia

### 1^2 + 2^2 + 3^2 + ... + n^2

Find a positive integer n > 1 such that

1^2 + 2^2 + 3^2 + ... + n^2

is a square number.

show that

9*T + 1

25*T + 3

49*T + 6

are triangular.

## Thursday, April 14, 2011

### 2..2^2 - 2..1^2

2^2 - 1^2 = 3
22^2 - 21^2 = 43
222^2 - 221^2 = 443
2222^2 - 2221^2 = 4443
22222^2 - 22221^2 = 44443
222222^2 - 222221^2 = 444443
2222222^2 - 2222221^2 = 4444443

### 2..3^2 - 2..2^2

3^2 - 2^2 = 5
23^2 - 22^2 = 45
223^2 - 222^2 = 445
2223^2 - 2222^2 = 4445
22223^2 - 22222^2 = 44445
222223^2 - 222222^2 = 444445
2222223^2 - 2222222^2 = 4444445

### 6^2 - 5^2, 56^2 - 55^2, 556^2 - 555^2

6^2 - 5^2 = 11

Base-2 = 1011 ...Base-3 = 102 ... Base-4 = 23 ... Base-5 = 21 ...
Base-6 = 15 ... Base-7 = 14 ... Base-8 = 13 ... Base-9 = 12 ... Base-11 = 10

56^2 - 55^2 = 111

Base-2 = 1101111 ... Base-3 = 11010 ... Base-4 = 1233 ... Base-5 = 421
Base-6 = 303 ... Base-7 = 216 ... Base-8 = 157 ... Base-9 = 133

556^2 - 555^2 = 1111

Base-2 = 10001010111 ... Base-3 = 1112011 ... Base-4 = 101113 ... Base-5 = 13421
Base-6 = 5051 ... Base-7 = 3145 ... Base-8 = 2127 ... Base-9 = 1464

5556^2 - 5555^2 = 11111

Base-2 = 10101101100111 ... Base-3 = 120020112 ... Base-4 = 2231213 ...
Base-5 = 323421 ... Base-6 = 123235 ... Base-7 = 44252 ... Base-8 = 25547 ...
Base-9 = 16215

55556^2 - 55555^2 = 111111

Base-2 = 11011001000000111 ... Base-3 = 12122102020 ... Base-4 = 123020013
Base-5 = 12023421 ... Base-6 = 2214223 ... Base-7 = 641640 ... Base-8 = 331007
Base-9 = 178366

555556^2 - 555555^2 = 1111111

Base-2 = 100001111010001000111 ... Base-3 = 2002110011021 ... Base-4 = 10033101013
Base-5 = 241023421 ... Base-6 = 35452011 ... Base-7 = 12305251 ... Base-8 = 4172107
Base-9 = 2073137

## Wednesday, April 13, 2011

### Square root of summed prime numbers

Let's take the prime numbers below 500

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, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311, 313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389, 397, 401, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 467, 479, 487, 491, 499

The aim: to identify all sets of N consecutive primes such that the square root of their sum is prime.

N = 2 to 9.

### Palindromic number 252

252 has the property that it becomes a perfect square when multiplied or divided by 7 :

252/7 = 36 = 6^2 ..... 252 * 7 = 1764 = 42^2

Could you find any other such palindromic numbers?

Interesting: 252 * 252 = 441 * 144 (palindromic equality)

## Tuesday, April 12, 2011

A semiprime is http://en.wikipedia.org/wiki/Semiprime

A semiprime is a natural number that is the product of two (not necessarily distinct) prime numbers.

Semiprimes (or biprimes): products of two primes http://oeis.org/A001358

Numbers which are not the sum of two semiprimes http://oeis.org/A072966

An interesting question: Is this sequence finite?

That is to say, can every natural number n > 33 be the sum of two semiprimes?

## Monday, April 11, 2011

### In what base is ab/ba an integer?

For example, in what base is 41 / 14 an integer?

Base-2 to Base-36 Conversions

## Saturday, April 9, 2011

### 1/e - 1/(ne) < (1 - 1/n)^n < 1/e - 1/(2ne).

Show that for any integer n > 1, we have

1/e - 1/(ne) < (1 - 1/n)^n < 1/e - 1/(2ne)

### m!n! = k!

Find integer m, n, k (all > 1) such that m!n! = k!

### 1/x + 1/y = 1/z

Find all triples (x, y, z) of positive integers such that

1/x + 1/y = 1/z

### Interesting palindromic numbers

12345654321 is an interesting palindromic number, since it is the square of another palindrome: 12345654321 = 111111^2

### x!*y! - x! - y!

Find all possible pair(s) (x, y) of nonnegative integers such that

x!*y! - x! - y!

is a perfect square.

## Friday, April 8, 2011

### 50 cents, 25 cents

You can pay 50 cents in exactly 50 different manners - depending on how many coins of each kind you use.

In how many manners can you pay 25 cents?

### Area of the six-pointed star

EDIT
Previously, I posted a 5-pointed star ... Sorry for the joke.

Find the area of the six-pointed star if all edges are of length s,
all acute angles are 60 degrees and all obtuse angles are 240 degrees.

### Spirals and the Golden Section

PART 4: THE MYTH OF THE NAUTILUS SHELL

## Thursday, April 7, 2011

### 1!*1 + 2!*2 + 3!*3 + ... + n!*n

1!*1 = 1
1!*1 + 2!*2 = 5
1!*1 + 2!*2 + 3!*3 = 23
1!*1 + 2!*2 + 3!*3 + 4!*4 = 119
1!*1 + 2!*2 + 3!*3 + 4!*4 + 5!*5 = 719
1!*1 + 2!*2 + 3!*3 + 4!*4 + 5!*5 + 6!*6 = 5039

for n = 1, 2, 3, 4, 5, 6.

Guess the general law, and prove your guess.

## Tuesday, April 5, 2011

### 2-digit primes and flipping primes

Flip Text Converter http://www.fliptext.org/

List of 2-digit prime numbers

2-digit prime numbers "ab"
To flip ab + a^b becomes = q^ɐ + qɐ

11: 11 + 1^1 = 12 (= 2^2 * 3)
13: 13 + 1^3 = 14 (= 2 * 7)
17: 17 + 1^7 = 18 (= 2 * 3^2)
19: 19 + 1^9 = 20 (= 2^2 * 5)
23: 23 + 2^3 = 31 (prime) flip it -> 3^2 + 32 = 41 (a prime number)
29: 29 + 2^9 = 541 (prime) flip it -> 6^2 + 62 = 98 (= 2 * 7^2)
31: 31 + 3^1 = 34 (= 2 * 17)
37: 37 + 3^7 = 2224 (= 2^4 * 139)
41: 41 + 4^1 = 45 (= 3^2 * 5)
43: 43 + 4^3 = 107 (prime) flip it -> 3^4 + 34 = 115 (= 5 * 23)
47: 47 + 4^7 = 16431 (= 3 * 5477)
53: 53 + 5^3 = 178 (= 2 * 89)
59: 59 + 5^9 = 1953184 (= 2^5 * 67 * 911)
61: 61 + 6^1 = 67 (prime) flip it -> 1^9 + 19 = 20 (= 2^2 * 5)
67: 67 + 6^7 = 280003 (= 19 * 14737)
71: 71 + 7^1 = 78 (= 2 * 3 *13)
73: 73 + 7^3 = 416 (= 2^5 * 13)
79: 79 + 7^9 = 40353686 (= 2 * 20176843)
83: 83 + 8^3 = 595 (= 5 * 7 * 17)
89: 89 + 8^9 = 134217817 (= 31 * 67 * 64621)
97: 97 + 9^7 = 4783066 (= 2 * 2391533)

### An integer in base 10 converted to base b

An integer in base 10 with distinct digits can be converted to base b by reversing
its digits. Find the lowest value of b.

### cracking a code in unsolved murder case

FBI asks for help cracking a code in unsolved murder case

## Monday, April 4, 2011

### Lost in a Forest

50 years ago, R. Bellman asked a remarkable minimization question that can be phrased as follows:

A hiker is lost in a forest whose shape and dimensions are precisely known to him.
What is the best path for him to follow to escape from the forest?

### The Anti-Tiger Mother Approach

Finland's Educational Success?
The Anti-Tiger Mother Approach

### Math/Maths Podcast 41

What makes a mathematician, and who should communicate mathematics?

## Sunday, April 3, 2011

### Fun Facts

The symbols + and –, referring to addition and subtraction, ﬁrst appeared in 1456 in an unpublished manuscript by the mathematician Johann Regiomontanus (a.k.a. Johann Müller). The plus symbol, as an abbreviation for the Latin et (and), was found earlier in a manuscript dated 1417; however, the downward stroke was not quite vertical.

In 1631, the multiplication symbol × was introduced by the English mathematician William Oughtred (1574–1660) in his book Keys to Mathematics, published in London. Incidentally, this Anglican minister is also famous for having invented the slide rule, which was used by generations of scientists and mathematicians. The slide rule’s doom in the mid-1970s, due to the pervasive inﬂux of inexpensive pocket calculators, was rapid and unexpected.

### Math Quotes #4

“Mathematics, rightly viewed, possesses not only truth, but supreme beauty — a beauty cold and austere, like that of sculpture”

(Bertrand Russell, Mysticism and Logic, 1918).

"In our private life as in our collective life there is no other truth than a statistical one."

-- Simone de Beauvoir

“We now know that there exist true propositions which we can never formally prove. What about propositions whose proofs require arguments beyond our capabilities? What about propositions whose proofs require millions of pages?
Or a million, million pages? Are there proofs that are possible, but beyondus?”

(Calvin Clawson, Mathematical Mysteries).

### A palindromic prime formed from the reflected decimal expansion of π

314 1592653589 7932384626 4338334626 4832397985 3562951413

### A good approximation of π

A good approximation of π is 43^(7/23)

## Friday, April 1, 2011

### Letters Triangle

The letters represent different natural numbers under 100

A
B    C
D    E    F
G    H    I    J
K    L    M    N    O
P    Q    R    S    T    U

Each letter is the sum of the two natural numbers just below it.

What is each letter if C > B?