Mathematics Fun,
Fact, Fiction,
Function, Fantasy





Here is a collection of mathematical activities and curiosities...enjoy them!

Guess Your Birthday!
 Divisibility Rules!
 Send More Money!
 Good Mathematical Card Trick
Did You Know?
 The Five Most Important Numbers in Mathematics in One Equation
 Trachtenberg System of Speed Mathematics
 Facts about Regular Polygons and Polyhedra
The Largest Known Prime Number 
Strange Properties of the Number 666
Beat The Calculator!
500 Digits of e 
 Amicable Numbers
 Largest Calculation of Pi
Magic Trick
 C Program for Calculating Pi
 The Magic Tetrahedron
 The Answer from the Ashes Magic Trick
  Infinite Integer Sequences
Interesting and Little-Known Algebra and Geometry Facts
 Do Ya Love the Number 17?
 Numbers Larger Than a Googolplex?
 The Music in the Numbers
 Chronological List of Mathematicians
 Incredible Human Calculators
 Newton's Method for Square Roots
 What Day of the Week Were You Born On?
 Interference of Sinusoidal Waveforms
The Encyclopedia of Polyhedra
Earliest Known Uses of Mathematical Terms
 Fermat's Last Theorem
 The Four 4's Puzzle
 Magic Squares
 Arithmetic Curiosities
Chisenbop Arithmetic
A Very Strange Prime Number
Towers of Hanoi
Towers of Brahma
"Handy" Multiplication
Numerology Your Interest?
Triangular Numbers
Whimsical Measurements Fascinating Method for Finding Pi Monty Hall Problem Simulation Let's Make a Deal (pdf) Experimental Economics Project
The Phenomenon of CoPrimes The Birthday Problem      

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Guess Your Birthday!

Here's a fun trick to show a friend, a group, or an entire class of people. I have used this fun mathematical trick on thousands of people since 1963 when I learned it. Tell the person (or class) to think of their birthday...and that you are going to guess it.

Step 1) Have them take the month number from their birthday: January = 1, Feb = 2 etc.
Step 2) Multiply that by 5
Step 3) Then add 6
Step 4) Then multiply that total by 4
Step 5) Then add 9
Step 6) Then multiply this total by 5 once again
Step 7) Finally, have them add to that total the day they were born on. If they were born on the 18th, they add 18, etc.

Have them give you the total. In your head, subtract 165, and you will have the month and day they were born on!

How It Works: Let M be the month number and D will be the day number. After the seven steps the expression for their calculation is:

5 (4 (5 M + 6 ) + 9 ) + D = 100 M + D + 165

Thus, if you subtract off the 165, what will remain will be the month in hundreds plus the day!

By the way, if you wish to know how many minutes and seconds you have been alive and lots more about when you were born, you might like to try this website ==> Birthday Calculator

Divisibility Rules!

To find if some number X is divisible by a certain number, test the number by using the information in the table below.

 By 2  If the last digit divisible by two, then X is too
 By 3  If the sum of the digits of the number X is divisible by three, then X is too
 By 4  If the last two digits are divisible by four, then X is too
 By 5  If the last digit is 5 or 0, then X is divisible by 5
 By 6  If X is divisible by 2 and by 3, then X is divisible by 6
 By 7  This rule is called L-2M. What you do is to double the last digit of the number X and subtract it from X without its last digit. For instance, if the number X you are testing is345678, you would subtract 16 from 34567. Repeat this procedure until you get a number that you know for sure is or is not divisible by seven. Then the X's divisibility will be the same.
 By 8  If the last three digits are divisible by 8, then X is too
 By 9  If the sum of the digits of the number X is divisible by nine, then X is too
 By 10  If the last digit of X is 0, then X is divisible by 10
 By 11  What you do here is to make two sums of digits and subtract them. The first sum is the sum of the first, third, fifth, seventh, etc. digits and the other sum is the sum of the second, fourth, sixth, eighth, etc. digits. If, when you subtract the sums from each other, the difference is divisible by 11, then the number X is too
 By 12  If X is divisible by 4 and by 3, then X is divisible by 12
 By 13  This rule is called L+4M. What you do is to quadruple the last digit of the number X and add it from X without its last digit. For instance, if the number X you are testing is345678, you would add 32 to 34567. Repeat this procedure until you get a number that you know for sure is or is not divisible by thirteen. Then the X's divisibility will be the same.
 By 14  If X is divisible by 7 and by 2, then X is divisible by 14
 By 15  If X is divisible by 5 and by 3, then X is divisible by 15
 By 16  If the last four digits are divisible by 16, then X is too
 By 17*  This rule is called L-5M. See rules for 7 and 13 on how to apply.
 By 18  If X is divisible by 9 and by 2, then X is divisible by 18
 By 19*  This rule is called L+2M. See rules for 7 and 13 on how to apply.
 By 20  If X is divisible by 5 and by 4, then X is divisible by 20
 By 21  If X is divisible by 7 and by 3, then X is divisible by 21
 By 22  If X is divisible by 11 and by 2, then X is divisible by 22
 By 24  If X is divisible by 8 and by 3, then X is divisible by 24
 By 25  If the last two digits of X are divisible by 25, then X is too
Higher  You can use multiple rules for multiple divisors...for instance, to check if a number is divisible by 57, check to see if it is divisible by 19 and 3, etc., since 57 = 19 x 3...

*A big thank you to Torsten Sillke for these rules!

Alphametics --- the alphabetic and mathematical construction where each letter is represented by a unique number in the problem. Each of these has a unique solution.


Do You Love "Alphametics?" Do You Want More? Go HERE! (Mike Keith's Great Site)

Good Mathematical Card Trick

Here's a real clean card trick that is bound to amaze and surprise your friends or classmates in your MBA programs.

Have a friend shuffle a standard 52-card deck to his satisfaction. The ask him/her to turn over, face up, a pile of twenty-five cards. As they count out the cards to twenty-five, act like you are intensely memorizing every single one of them in order. In truth you are only interested in the 17th card in the pile. Memorize that 17th card..

Turn the twenty-five card pile over, now face down, and set aside.

Take the remaining cards and do the following:

If it is a two through nine card, place it face up, and count out cards up to the number ten. For instance, if you turn up a six, you would place it face up, and then count out four cards coming down, saying "Seven, eight, nine, ten." If you turned up a three, then seven cards, saying "Four, five, six, seven, eight, nine, ten."

If the card you turn up is an ace, ten, or picture, tell your friend that you must have cards 2 through nine for this trick and place the card on the bottom of the deck, face down like the rest of the cards. (Someone told me that aces and pictures work too, but I haven't verified it yet.)

Repeat this procedure until you have completed four columns of "ten counts."

Take whatever remaining cards you have after making the four columns, and place them face down on TOP of the twenty-five card pile you set aside earlier.

Now ask your friend to total up the numbers at the top of the four columns.

Say the total was 23. Counting from the very top of the set-aside-pile-plus-remainder-cards, state you are interested in the 23rd card. Just before turning over the 23rd card, state what card it is!! (Iit is the 17th card that you memorized at the beginning of the trick!)

If you did everything correctly, it works every time! Amazing!

Did You Know?

Did you know that 27^5 + 84^5 + 110^5 + 133^5 = 144^5 ?

Did you know that 9^3 + 10^3 = 12^3 + 1^3 ?

Did you know that 2682440^4 + 15365639^4 + 18796760^4 = 20615673^4 ?

Did you know that 95800^4 + 217519^4 + 414560^4 = 422481^4 ?

Did you know that 111,111,111 x 111,111,111 = 12,345,678,987,654,321 ?

Did you know that 123,456,789 x 989,010,989 = 122,100,120,987,654,321 ?

Did you know that e^(pi * sqrt(163)) = 262537412640768743.9999999999992 ? Coincidence?

Did you know there are just four numbers (after 1) which are the sums of the cubes of their digits:
153 = 1^3 + 5^3 + 3^3
370 = 3^3 + 7^3 + 0^3
371 = 3^3 + 7^3 + 1^3 and
407 = 4^3 + 0^3 + 7^3 ?

Did you know 1,741,725 = 1^7 + 7^7 + 4^7 + 1^7 + 7^7 + 2^7 + 5^7 ?

Trachtenberg System of Speed Mathematics

When I was just seven or eight years old, I came upon a most fascinating book called the Trachtenberg System of Speed Mathematics (or something like that). In it was described a way of doing addition, subtraction, multiplication and division in ways I had never seen before. None of my teachers had even heard of it before. I was able to do multiplications by eleven and twelve in my head faster than my friends could do on paper. If you hate formal math (or if you have a child that does), I urge you to download the software that lays it all out for you.

Download the software directly from this site.

If the link is down, email me and I will send it along to you.

Two other sites which offer it are here and here.

Jakow Trachtenberg was born on June 17, 1888, in Odessa, Russia. He worked as an engineer during his younger years in the Obuschoff Shipyards, after graduating with highest honors from the Mining Engineering Institute in St. Petersburg. He moved to Berlin during World War I and soon became an expert in Russian affairs. There he devised a method of teaching languages that is still used in some parts of eastern Europe. As a Jew, he was captured by the Nazis and deported to a concentration camp. There, without paper or pencil (some say a common iron nail became his most prized possession), he devised a method of doing mental arithmetic. Trachtenberg managed to escape and fled to Switzerland, where he perfected his system. In 1950, he founded the Mathematical Institute in Zurich where he taught his methods to children and adults alike.

Facts about Regular Polygons and Polyhedra

The following legend pertains to the table below it:

 s = length of a side  A = area  d = diagonal
 R = radius of circumscribed circle or sphere  V = volume  a = length of an edge
 r = radius of inscribed circle  h = height  S = surface area

 Triangle  A = ((s^2)*sqrt(3)) / 4
Seven ways to find the area of a triangle...go
 Tetrahedron  V = (1/12) * sqrt(2) * a^3
 R = (1/3) * s * sqrt(3)  S = a^3 * sqrt(3)
 r = (1/6) * s * sqrt(3)  R = (1/4) * a * sqrt(6)
Isosceles Area = (c/4)(sqrt(4a^2 - c^2)
where (c is the base, a are the legs)
 r = (1/12) * a * sqrt(6)
 Square  A = s^2  Cube  V = a^3
 R = (1/2) * s * sqrt(2)  S = 6 * a^2
 r = (1/2) * s  R = (1/2) * a * sqrt(3)
   r = (1/2) * a
 Pentagon  A = (s^2 / 4) ( sqrt( 25 + 10 * sqrt(5) ) )  Octahedron  V = (1/3) * a^3 * sqrt(2)
 R = (1/10) * s * sqrt (50 + 10 * sqrt(5) )  S = 2 * a^2 * sqrt(3)
 r = (1/10) * s * sqrt (25 + 10 * sqrt(5) )  R = (1/2) * a * sqrt(2)
   r = (1/6) * a * sqrt(6)
 Hexagon  A = (3/2) * s^2 * sqrt(3)  Dodecahedron  V = (1/4) * a^3 * (15 + 7 * sqrt(5) )
 R = s  S = 3 * a^2 * sqrt(25 + 10 * sqrt(5) )
 r = (1/2) * s * sqrt(3)  R = (1/4) * a * ( sqrt(3) + sqrt(15) )
   r = (1/4) * a * sqrt( (50 + 22 * sqrt(5)) / 5 )
 Octagon  A = 2 * s^2 * (1 + sqrt(2) )  Icosahedron  V = (5/12) * a^3 * (3 + sqrt (5) )
 R = (1/2) * a * sqrt( 4 + 2 * sqrt(2) )  S = 5 * a^2 * sqrt(3)
 r = (1/2) * a * (1 + sqrt(2) )  R = (1/4) * a * sqrt(10 + 2 * sqrt(5) )
   r = (1/2) * a * sqrt( (7 + 3 * sqrt (5)) / 6 )
 Decagon  A = (5/2) * s^2 * sqrt (5 + 2 * sqrt(5))  Sphere  V = (4/3) * pi * a^3
 R = (1/2) * a * (1 + sqrt(5) )  S = 4 * pi * a^2
 r = (1/2) * a * sqrt( 5 + 2 * sqrt(5) )  
 Cone S = (pi) * r * sqrt(r^2 + h^2)  Radius of circle inscribed in a triangle  R = sqrt[s * (s - a) * (s - b) * (s - c)] / s
where s = (1/2) * (a + b + c)

Magic Trick --- Take any whole number greater than 0. Take half of it if it is even or triple it and add one if it is odd. Repeat over and over again. You'll always get the same result ---> 1.

The Five Most Important Numbers in Mathematics in One Equation

e^i*pi + 1 = 0

By the way, you can see the value of e to 10,000 digits HERE! (Thanks to the University of Utah)

Infinite Integer Sequences? There's a whole encyclopedia devoted to them! HERE!

Primes! As of 2003, the largest known prime number has been verified as 2^13466917 - 1 . Known as the 39th Mersenne prime number, that is, of the form 2^p - 1, where p is also prime, it has 4,053,946 whopping digits! You can see it HERE!

In February, 2005, an even larger Mersenne prime was found! Yes, believe it or not, the 42nd Mersenne prime was independently verified as 2^25,964,951-1, an astounding number having 7,816,230 incredible digits...there is a poster available of it too...I'll have to find where...

WAIT!! - NEWS FLASH - At UCLA, the 46th Mersenne prime was just discovered! Its incredible value has over 13 million digits!! The number is 2^43,112,609 - 1. They won a $100,000 prize from the Electronic Frontier Foundation for finding it. It is the eighth Mersenne prime discovered at UCLA.

By the way, they are named for the 17th century French mathematician Marin Mersenne, who discovered them.

For more info on Mersenne primes, go HERE!


As of February, 2013 -->  LARGEST PRIME NUMBER FOUND (48th Mersenne): Get ready for some even tougher answers on
those math tests. Mathematicians—with the aid of a giant network of computers—said on Tuesday that they had discovered the largest prime
number—and it’s 17,425,170 digits long. Discovered by University of Central Missouri mathematician Curtis Cooper, the number—2 raised to
the 57,885,161 power minus 1—crushes the last prime number discovered, which had only 12,978,189 digits. “It’s analogous to climbing Mount
Everest,” said retired computer scientist George Woltman. The number is also the 48th of a rare group of primes known as the Mersenne Primes,
which take the form of 2 raised to the power of a prime number minus 1—first described by French monk Mersenne 350 years ago...there's an
article about it HERE.

What a Slice of Pi! As of 2005, the greatest calculation of pi has been done by Professor Yasumasa Kanada and a team of researchers who set a new world record by calculating the value of pi to 1.24 trillion places (that's 1,240,000,000,000). The previous record, set by Kanada in 1999, was 206.158 billion about it HERE! For the first 100,000 digits of pi, go HERE! You can even search for your birthday in pi! Try HERE. Some guy has even memorized the first 40,000 digits of about him and lots of other amazing memory feats HERE!

We can also conjure pi out of the whole numbers, like this:
pi^2/6 = 1/1^2 + 1/2^2 + 1/3^2 + 1/4^2 + 1/5^2 + . . .
Or we can do it with just the odd numbers:
pi/4 = 1 – 1/3 + 1/5 – 1/7 + 1/9 – 1/11 + . . .
Or their squares:
pi^2/8 = 1/1^2 + 1/3^2 + 1/5^2 + 1/7^2 + 1/9^2 + . . .
Or with evens and odds combined:
pi = 2/1 * 2/3 * 4/3 * 4/5 * 6/5 * 6/7 * . . .

(Thanks to Toby Howard)


C Program for Calculating Pi --- Here is a 160 character program in C, written by D. T. Winter, which will calculate the first 800 digits of pi:

int a=10000,b,c=2800,d,e,f[2801],g;main(){for(;b-c;)f[b++]=a/5;



Strange Properties of 666, "The Number of the Beast"

The last book of the Bible, Revelation, brings up the number 666 as being the number of the beast connected with the end of this age and the coming of the Messiah. You will find the direct reference in Chapter 13, verse 18 of Revelation. Besides that cataclysmic reference, the number 666 has quite a few very interesting properties.

666 = 3^6 - 2^6 + 1^6

666 = 6^3 + 6^3 + 6^3 + 6 + 6 + 6

(Mike Keith mentions that there are only five other positive integers that exhibit this property...find 'em!)

666 = 2^2 + 3^2 + 5^2 + 7^2 + 11^2 + 13^2 + 17^2

666 = 1 + 2 + 3 + 4 + 567 + 89 = 123 + 456 + 78 + 9 = 9 + 87 + 6 + 543 + 21

Moreover, 666 is equal to the sum of the cubes of the digits in its square (666^2 = 443556, and the sum of the cubes of these digits is 4^3 + 4^3 + 3^3 + 5^3 + 5^3 + 6^3 = 621) plus the sum of the digits in its cube (666^3 = 295408296, and 2+9+5+4+0+8+2+9+6 = 45, and 621+45 = 666).

Incredibly, the number 666 is equal to the sum of the digits of its 47th power, and is also equal to the sum of the digits of its 51st power. That is,

666^47 = 5049969684420796753173148798405564772941516295265

666^51 = 9935407575913859403342635113412959807238586374694

and the sum of the digits on the right hand side is, in both cases, 666. In fact, 666 is the only integer greater than one with this property. (Also, note that from the two powers, 47 and 51, we get (4+7)(5+1) = 66.)

Mr. Keith also points out that if we assign numerical values for the letters of the alphabet starting with A = 36, B = 37, and so on, we find that the letters in the word


From contributor James Watt comes the following: "Here are some other neat things about 666 I seem to have
discovered (since I never found any reference anywhere else). 6+6+6 =18 and 18 x 37 =
Similarly, 4+4+4 = 12. 12 x 37 = 444. etc. In Roman numerals (and the Greek equivalents), which John would have used
to write them, it is DCLXVI, the exact sequential descending order of Roman Numerals.
1/81 = .012345679012345679012345679.... Notice the 8 is missing. 1+2+3+4+5+6+7+9 = 37.
The other 'number of the beast' is called the vulgate number. It is 616. If 'vulgar' Roman numerals are used in
ascending order, vulgate Roman numerals 616 is IVXCD. The'L' is missing."

From contributor Matt Westwood comes this interesting fact: "666 is also the total of all the numbers in a 6x6 magic square. That is, it's the 36th triangle number, being <math>sum_{k=1}^{36} k</math>." He also goes on to mention "In one of the many branches of Hermitic magic that I once studied, the magic squares of the various sizes were talismans for the various planetary influences: 3x3: Saturn, 4x4: Jupiter, 5x5: Mars, 6x6: Sun, 7x7: Venus, 8x8: Mercury, 9x9: Moon. A more recent book on the subject tried to convince that 11x11 was Neptune, or something, but this is far from profound: there exists a simple algorithm for creating *any* odd-order magic square with consecutive natural numbers starting from 1. Although I did find it profound at the time that I won a raffle ticket with the number 369 (the number of the 9x9 square) on the very day I moved into a new house. Add or subtract n from every digit of any magic square, and it remains magic of course."

15) The Answer from the Ashes Magic Trick

This is my all-time favorite mathematical magic trick. Choose a group setting like a big holiday family dinner, an informal teacher conference, a math class, or even a party. You'll need a pencil, two notebook size pieces of paper, a large empty ashtray, some matches or lighter, and a bar of soap.

Preparation: Ten minutes before you begin the trick, excuse yourself and go to the bathroom. There take a bar of soap in hand and with the corner of the moistened end, carefully draw the number 1089 on top of your left forearm. Let it air dry.

When you begin this trick, ask for someone in your audience who believes they have ESP ability and can do addition and subtraction correctly. What's even better is to ask that rich uncle or grandfather to do the trick with you!

Step 1) Tell them to choose any three digit number where the first and last digits differ by two or more. Write it down on the first piece of paper. Numbers like 187, 249, 386 are fine, but number like 172, 584, and 928 are not.

Step 2) Tell them to reverse the digits and subtract the smaller number from the larger. For instance, if they initially chose 672 as their number, they would be subtracting 276 from the 672 to get 396.

Step 3) Then tell them to reverse the digits of the difference they found and now add them together. In our example, you would add 693 to the 396. (Of course, notice that the total will be 1089...big surprise, huh?)

Step 4) Tell them to double check their work and to show everyone the final result as you turn your back. Then tell them to write the final result nice and big on the second piece of paper.

Step 5) Tell them to fold up both pieces of paper three or four times so that you could not see what was written on them.

Step 6) Put both pieces of paper in the large ashtray, ignite, and burn completely into ashes.

Step 7) Tell your mark to concentrate intensely on the final result number and tell your audience that you will divine the number directly from the ashes. Go into your Kreskin act. Then pick up some ashes and rub them directly onto your soap-written arm. As you continue to rub the ashes, they will stick to the dried soap and incredibly the answer
1089 will appear on your arm!!

Beat Your Calculator! --- If you are one of those people who feel it is somewhat self-demeaning to use a calculator to do simple two-digit multiplications or other simple mathematical tasks, this page is for you! Go HERE!

17 Page --- For those of you who love the number 17, there is an entire web page devoted to it. HERE!

Numbers Larger Than a Googolplex?--- Well, yes, there are! Quite a few actually...but let me let an expert describe Graham's Number, Knuth's Notation, Ackerman's Function, Conway's Notation, and Moser's Number: HERE!

The Music in the Numbers --- Did you know that fractal music be produced using pure numbers? Lars Kindermann of the University of Erlangen has produced MusiNum, a program for PCs that does just that. Go HERE!

The Chronological List of Mathematicians --- There is a listing of every famous, near-famous, and not-so-famous mathematician that ever lived...starting around 1650 B.C. to the present...check it out HERE!

Newton's Method for Calculating a Square Root --- This method involves a bit of multiplying and dividing but it will arrive at the precise square root over time and trials:

Step 1) Let X be the number of which you wish to find the square root. Let G be your best guess at each stage of the calculation.
Step 2) To find the next G, call it G', you set G' equal to:

G' = ( X + G^2) / 2G

Step 3) This G' becomes your next G. Iterate, that is repeat the calculation over and over (as shown below), and you will eventually arrive at the square root you seek.

G'' = ( X + G'^2) / 2G'

G''' = ( X + G''^2) / 2G''

If you're interested in methods you can employ to find cube roots, go HERE or HERE!

What Day of the Week Were You Born On? --- Even though you were there at the moment of your birth, you may not remember exactly what day of the week it was. In fact, not only will this method help you find that out, you can find out the day of the week for any date you want in the 1900's. Here is a little trick to help you figure what day that was:

Step 1) Write the last two digits of the year you were born. Call it A.
Step 2) Divide that number, that is, divide A by four. Drop the remainder if there is one.
Call this answer, without the remainder,
Step 3) Find the month number corresponding to the month you were born in from the table below. Call it
Step 4) Oh, the date you were born on, call it
D. (If you were born on the 12th, call D 12.)
Step 5) Now add
A + B + C + D. Divide this sum by 7. The remainder you get is the key to the day of the week.
Step 6) In the table of days below, match the remainder with the day of the week you were born on.
NOTE: This trick will work for any date in the 20th century.

       Sunday = 1
 January = 1 (0 in leap yr)  July = 0    Monday = 2
 February = 4 (3 in leap yr)  August = 3    Tuesday = 3
 March = 4  September = 6    Wednesday = 4
 April = 0  October = 1    Thursday = 5
 May = 2  November = 4    Friday = 6
 June = 5  December = 6    Saturday = 0

From Arthur Overton comes an enhanced version of this that will allow you to seek out the day of the week for ANY A.D. year, not just 20th century ones...thank you kindly Arthur!

Step 1) Write down the last two digits of the year. Call it A.
Step 2) Divide that number by four dropping the remainder call this answer
Step 3) Find the month number corresponding to the month from the table below. Call it
Step 4) The day call it
Step 5) Take the century number of the year and divide it by four.
If the remainder is 0 then E = 6
If the remainder is 1 then E = 4
If the remainder is 2 then E = 2
If the remainder is 3 then E = 0
Step 6) Now add
A+B+C+D+E. Divide this sum by 7. The remainder you get is the key to the day of the week.
Step 7) In the table of days below, match the remainder with the day of week.

    Sunday = 1   If year divides 400, yes
Jan = 1 (0 in leap year) July = 0 Monday = 2   If year divides 100, no
Feb = 4 (3 in leap year) Aug = 3 Tuesday = 3   If year divides 4, yes
Mar = 4 Sept = 6 Wednesday = 4   If year does not divide 4, no
April = 0 Oct = 1 Thursday = 5    
May = 2 Nov = 4 Friday = 6    
June = 5 Dec = 6 Saturday = 0    

**From Alison comes this comment, "In the enhanced Overton method, for determining which day of the week based on birthdate, it seems the table relating the remainder after division by 7 with the days of the week is offset by one day. I believe it should read Sunday = 0, Monday = 1, Tuesday = 2, etc." THANK YOU, ALISON!

Amicable Numbers --- There are a few pair of numbers that have a very peculiar affinity for each other and are so-called "amicable numbers." Take for instance the pair of numbers 220 and 284. It turns out that all the factors of 220, that is those less than itself, add up to 284. And, surprisingly, the factors of 284 add up to 220. I only know of three other pairs like these: 1,184 and 1,210 (discovered by a 16-year-old Italian named Nicolo Paganini), 17,296 and 18,416, and the large pair 9,363,584 and 9,437,056. Can you find others?

Incredible Human Calculators --- Oh you might know of someone who can do two- or even three-place multiplication problems in their head, and there are those who can add faster than you can using an electronic calculator, but do you know the stories of Zerah Colburn and Truman Henry Safford?

Zerah Colburn was born in 1804, the son of a Vermont farmer. By the age of eight, he was giving mathematical exhibitions in England where he was asked by a member of the audience to compute 8 to the 16th power. He gave the correct answer 281,474,976,710,656 in about thirty seconds, and brought the astounded audience to tears. Zerah eventually stayed in England, received his formal education there, but strangely his incredible calculating abilities waned as he aged. He died in 1840, at the age of just 36, after a life of teaching Greek, Latin, French, Spanish and English in the United States, but not before writing his autobiography in which he outlined his calculating methods.

Another calculating prodigy, Truman Henry Safford, was born in 1836, coincidentally in Vermont. When he was ten, he was given a problem in church by the Reverend H. W. Adam: Multiply in your head 365,365,365,365,365,365 by itself! According to the good reverend's own account, Truman "flew around the room like a top, pulling his pantaloons over the tops of his boots, biting his hands, rolling his eyes in their sockets, sometimes smiling and talking, and then seeming to be in agony." In less than one minute, though, he had come up with the correct answer: 133,491,850,208,566,925,016,658,299,941,583,225!!! The boy admitted he was exhausted after this calculation. He never did any public exhibitions, went to college, studied astronomy, and like Zerah Colburn, lost much of his amazing abilities as he aged. He died in 1901.

There have been other famous and not-so-famous child prodigies such as John Wallis, the great Johann Carl Friedrich Gauss, Andr Marie Ampere, George Parker Bidder (Senior and Junior), Johann Martin, Zacharias Dase, Jacques Inaudi, and Shakuntale Devi. A nice list and description of other mathematical child prodigies can be found HERE.

Interesting and Little-Known Algebra and Geometry Facts --- Here are a few helpful and neat little facts that evade most students and teachers of algebra and geometry:

 a) Though everyone can factor a^2 - b^2, a^3 - b^3, a^3 + b^3 and a^4 - b^4, most folks do not know that:
 a^4 + b^4 = (a^2 + ab(sqrt(2)) + b^2) (a^2 - ab(sqrt(2)) + b^2
 b) An approximation exists for the factorial function (for large n) which seems hardly related but works:
Stirling's Formula --->  n! ~~ e^(-n) n^n sqrt(2 (pi) n)
 c) The area of any regular polygon of n sides, each of length x, is given by:
 Area = (1/4)nx^2 (cot(180/n))
 d) The radii of circumscribed (R) and inscribed (r) circles within such regular polygons are given by:
 R = (x/2) csc (180/n) and r = (x/2) cot (180/n)
 e) The radius of a circle inscribed within any triangle of sides a, b, and c with semi-perimeter s is given by:
r = (sqrt (s (s-a) (s-b) (s-c)) / s 
 e) The radius of a circle circumscribed about any triangle of sides a, b, and c with semi-perimeter s is given by:
R = abc / 4 (sqrt (s (s-a) (s-b) (s-c))
 f) The perimeter P and area A of polygons (of n sides) inscribed in a circle of radius r is given by:
 P = 2nr sin(pi/n) and A = (1/2) nr^2 sin (2pi/n)
 g) The perimeter P and area A of polygons (of n sides) circumscribed about a circle of radius r is given by:
 P = 2nr tan (pi/n) and A = nr^2 tan (pi/n)
h) Factoring the seemingly prime expression a^4 + 4b^4 becomes (and there are a family of these)
a^4 + 4b^4 = (a^2 + 2b^2)^2 - (2ab)^2 (thanks to N. Hobson)

The Magic Tetrahedron (many thanks to R. Leo Gillis) --- Here's a neat trick involving all the numbers from 1 to 26, and three of the five Platonic Solids, the most basic polyhedral shapes. Let's start with the tetrahedron. A tetrahedron, sometimes called a triangular pyramid, is a shape made up of four corners, four equilateral triangular faces, and six edges. Since each face is a triangle, it also has a total of 12 angles on its four faces. So the tetrahedron has a total of 26 components, (4 corners + 4 faces + 6 edges + 12 angles = 26). These components can be numbered from 1 to 26 in a special way.

The basis of the trick is to use three pairs of numbers in order to create a value for every part of the tetrahedron. The three pairs are: 1 & 2, 3 & 6, 9 & 18. These six numbers will be placed on the six edges of the tetrahedron. Each edge is always directly opposite another edge; that is, if you draw a line through the center of the object starting from one edge, you will always reach another edge on the opposite side.

Select any edge and place the number 1 on it. On the opposite edge place the number 2. Select any of the remaining edges and place the number 3 on it, and then place the number 6 on the opposite edge. On the last two edges place the numbers 9 and 18. Now you're ready to determine all the rest of the numbers, and where they go on the tetrahedron.

Every corner has three edges that meet at it. Add up the value of the three edges and give that number to the corner. Every side has three edges surrounding it. Add up the value of the three edges and give that number to the face. Every angle on the faces is formed by two edges meeting there. Add the value of these two edges and give that number to the angle.

When you are done, you will discover that you have used all the numbers from 1 to 26 without repeating any numbers! There are two possible tetrahedra that can be made this way. Can you find them both?

This trick can also be done on a cube, but without using the angles of the faces. A cube has six faces, eight corners, and twelve edges; 6 + 8 + 12 = 26. Using the same three pairs of numbers we started with above, we can give values to the six faces. Place the number 1 on a face, and the number 2 on the opposite face. Place the number 3 on any of the remaining faces, and the number 6 on the opposite face. Then place the numbers 9 and 18 on the remaining faces.

Just like before, we will add these six numbers together to create all the others. Every edge of a cube is the meeting place of two faces. Add the numbers of these two faces and give that total to the edge. Every corner of a cube is the meeting place of three faces. Add the numbers of these faces together and give that total to the corner. Do this for all parts of the cube, and you will use all the numbers from 1 to 26 without any repetitions. Unlike the tetrahedron, there is only one way to accomplish this feat on a cube.

The technique will also work on the octahedron, which has six corners, eight sides, and twelve edges; 6 + 8 + 12 = 26. In this case the three pairs of opposite numbers are placed on the six corners of the octahedron. As before, place 1 and 2 on opposite corners, 3 and 6 on opposite corners, and 9 and 18 on the remaining corners. Each of the twelve edges of the octahedron connects two corners. Add these two corners to find the number for that edge. Each of the eight triangular sides of the octahedron is surrounded by three corners. Add those three corners together to get the value for that side. When completed, you will again have used all the numbers from 1 to 26 with no repetitions. As with the cube, there is only one way to accomplish this.

The sharp observer will have noticed that the special set of 3 opposite pairs are used for the edges of a tetrahedron, the faces of a cube, and the corners of an octahedron, which may reveal something important about the relationships of points in space. Also, for the die-hard puzzle solver, the solutions to the above figures may be written in Base 3 numerals, with very interesting results.

500 Digits of e - Named after the world famous mathematician and extreme child prodigy Leonhard Euler, the natural logarithmic base has innumerable applications in all fields of science, business, and is just the first 500 digits or so...

2.71828 18284 59045 23536 02874 71352 66249
77572 47093 69995 95749 66967 62772 40766 30353 54759 45713
82178 52516 64274 27466 39193 20030 59921 81741 35966 29043
57290 03342 95260 59563 07381 32328 62794 34907 63233 82988
07531 95251 01901 15738 34187 93070 21540 89149 93488 41675
09244 76146 06680 82264 80016 84774 11853 74234 54424 37107
53907 77449 92069 55170 27618 38606 26133 13845 83000 75204
49338 26560 29760 67371 13200 70932 87091 27443 74704 72306
96977 20931 01416 92836 81902 55151 08657 46377 21112 52389
78442 50569 53696 77078 54499 69967 94686 44549 05987 93163
68892 30098 79312 77361 78215 42499 92295 76351 48220 82698
95193 66803 31825 28869 39849 64651 05820 93923 98294 88793
32036 25094 43117 30123 81970 68416 14039 70198 37679 32068
32823 76464 80429 53118 02328 78250 98194 55815 30175 67173

Interference of Sinusoidal Waveforms Etc. - There is a site that will graphically show you how two sinusoids interact, be they in phase or our of phase, be they the same amplitude or different, or being the same wavelength or different...very instructive...go see it HERE. The same teacher that introduced me to this page also has a page to explain "damped" functions, a very important phenomena in virtually every engineering can see it HERE.

Fermat's Last Theorem - The Proof - Here is a link to PBS' Nova On Line website that describes the 350-year search for the proof of one of mathematics' greatest can find it HERE.

The Four 4's Puzzle - When I was teaching SAT math, I had occasion to try my hand at this puzzle...the idea is that by using four 4's and any combination of mathematical operations, one can write expressions that equal the values from 0 to 100 as the answer. Now I finished it and my notes are somewhere in my classroom file cabinet, but I found a place where a solution is HERE you go...try your own hand at it before cheating! It's more fun that way!

Magic Squares - There are a number of websites dedicated to these little know, the square grids of numbers where the rows, columns, and diagonals all add up to the same number...well, you'll be impressed how far a little concept like that can take you...try HERE, or maybe HERE...even HERE for a ton of magic square links...even find their history is HERE...

The Encyclopedia of Polyhedra - Though I include some very interesting polyhedra texture-mapped with fractals in some of my fractal compositions, and even show a few on separate gallery pages, I hardly give them the extremely comprehensive attention that George Hart does in his very impressive website...go will find thousands, yes thousands of virtual reality polyhedra to explore and enjoy...lots of very interesting mathematical fun for sure...

Chisenbop Arithmetic - Many years ago, I encountered a young Korean student in my class who showed me a set of really neat ways of doing arithmetic using his fingers...multiplying, adding, really quite intriguing, and so simple that, well, a child could learn wasn't until maybe ten years later that I heard of it is called Chisenbop...believe or not, there is even a website dedicated to it...HERE it is...

Chisenbop Multiplying by 9

Hold out your hands in front of you so that your thumbs point toward one another.

Visualize that your left pinky finger represents 1, the next finger 2, and so on left to right, until your right pinky finger represents 10. Those fingers represent the number you wish to multiply by 9. To do so, simply put the finger down you wish to multiply by 9. All fingers to the left of the down finger represent the tens digit of the answer while all fingers to the right represent the ones digit.

Example: 6 x 9. Put the finger representing 6 down (the right hand thumb). To the left of the down finger, you have 5 fingers up. That's your tens digit, 5. To the right, you have 4 fingers up. There's your ones digit, 4. Put those together and you have your answer: 54. Pretty cute.

Arithmetic Curiosities - Here are just a few interesting patterns in arithmetic that you or your students may explore. Verify these results with paper and pencil or with calculator (if you must):

 1 x 9 + 2 = 11 9 x 9 + 7 = 88 9 x 9 = 81 6 x 7 = 42
12 x 9 + 3 = 111 98 x 9 + 6 = 888 99 x 99 = 9801 66 x 67 = 4422
123 x 9 + 4 = 1111 987 x 9 + 5 = 8888 999 x 999 = 998001 666 x 667 = 444222
1234 x 9 + 5 = 11111  9876 x 9 + 4 = 88888 9999 x 9999 = 99980001  6666 x 6667 = 44442222

Earliest Known Uses of Some of the Words of Mathematics - HERE is a site that says it all!

A Strange Prime Number - The prime number 73,939,133 has a very strange property. If you keep removing a digit from the right hand end of the number, each of the remaining numbers is also prime. It's the largest number known with this property. Take a look: 73,939,133 and 73,939,13 and 73,939,1 and 73,939 and 7,393 and 739 and 73 and 7 are all prime! (Thanks to Toby Howard)

Towers of Hanoi - Tower of Brahma - The Towers of Hanoi Puzzle was invented in 1883 by Edouard Lucas, a French mathematician. This puzzle was originally sold as a toy, and was described as a simplified version of the "Tower of Brahma."

The legend has it that the mythical Tower of Brahma, in the Indian city of Benares, has a post with 64 gold discs stacked in decreasing sizes. These discs are said to be moved, one at a time, by the priests of the Hindu temple from the original post to one of the other two posts, with a larger disc never being placed on top of a smaller disc. It was said that by the time all 64 discs have been transferred to another post in the original order, the temple will crumble to dust and the earth would disappear.

At the rate of moving 1 disc per second around the clock, it is estimated to take about 585 billion years to complete the transfer of the gold discs in the Tower of Brahma. This was calculated using the formula: X = 2^n - 1, where X is the number of moves required and where n is the number of discs to be moved.

"Handy" Multiplication - Just when I thought I knew everything (lol), someone came along a showed me a very "handy" way to memorize and learn your times table, intended for the sixes thru the tens...

What you do is this: Hold your hands up in front of you, palms toward you. Number each pinkie 6, ring finger 7, tall man 8, pointer 9, and thumb 10...sounds silly I know, but it's pretty cool...

Let's say you wish to multiply 9 x 7...okay, touch the ring finger (the 7) on one hand to the pointer (the 9) of the other hand, and make a horizontal line, joining them ask, "How many finger are on this line or below?" Well there are six. That is your answer's tens digit. Now look at the fingers above the line...multiply them...1 times 3 is 3...that's your ones digit...thus 9 x 7 = 63...pretty cool , huh?

Numerology- Here is one of the most extensive websites devoted to numerology I have ever seen ==> The Number Wizard

Triangular Numbers - One interesting group of numbers are called triangular numbers. These numbers are those "that can be represented in the form of a triangular grid of points where the first row contains a single element and each subsequent row contains one more element than the previous one." A complete discussion of them can be found HERE.

Nick Hobson points out that you can "Take any triangular number and append the digit 1. For example, the 100th triangular number is 5050, so we get 50501. Then, all the divisors of the resulting number end in 1 or 9. At first sight it would seem a divisor could end in 3 or 7, but this never happens! For example, 50501 has divisors 1, 11, 4591, and 50501. See the solution to my puzzle 149 ( for further explanation."

He also points out that 666 (see the discussion about THAT number above) is the largest triangular number consisting of the same digit.

Whimsical Measurements - Now I know these have been going around the internet for many years...recently someone sent me a goodly-sized group of these playful little conversion factors...some of which are quite clever...see what you think...

* 4 lawyers = 2 paralegals
* 2 untruths = 1 paralyze
* All perjured testimony in a gov't coverup = 1 scandalize
* 33.8 oz of a case of soft drinks = 1 liter of the pack
* 1 first-date kiss = 1 peck
* 1 french kiss = two tongues of fun
* Ratio of an igloo's circumference to its diameter = 1 Eskimo Pi
* 2000 pounds of Chinese soup = Won ton
* 1 millionth mouthwash = 1 microscope
* Speed of a tortoise breaking the sound barrier = Mach Turtle
* Time it takes to sail 220 yards at 1 nautical mile per hour = Knot-furlong
* 365.25 days of drinking low-calorie beer because it's less filling = 1 lite year

* 16.5 feet in the Twilight Zone = 1 Rod Serling
* Half of a large intestine = 1 semicolon
* 1000 aches = 1 megahurtz
* Weight an evangelist carries with God = 1 billigram
* Basic unit of laryngitis = 1 hoarsepower
* Shortest distance between two jokes = A straight line
* Time between slipping on a peel and smacking the pavement =
1 bananosecond
* A half bathroom = 1 demijohn
* 453.6 graham crackers = 1 pound cake
* The first step of a one-mile journey = 1 Milwaukee
* 1 million microphones = 1 megaphone
* 1 million bicycles = 2 megacycles
* 365.25 days = 1 unicycle
* 2000 mockingbirds = 2 kilomockingbirds
* 10 cards = 1 decacards
* 1000 grams of wet socks = 1 literhosen
* 1 millionth of a fish = 1 microfiche
* 1 trillion pins = 1 terrapin
* 1 trillion bulls = 1 terrabull
* 1 million billion picolos = 1 gigolo
* 10 rations = 1 decoration
* 100 rations = 1 C-ration
* 10 millipedes = 1 centipede
* 3 1/3 tridents = 1 decadent
* 10 monologs = 5 dialogs
* 5 dialogs = 1 decalog
* 2 monograms = 1 diagram
* 4 nickels = 2 paradigms
* 2 wharves = 1 paradox
* 100 Senators = Not 1 decision

Fascinating Method for Finding Pi - From Jonas Castillo Toloza of Colombia comes this interesting method for finding pi...he has sent me numerous mathematical curiosities in the past...I just haven't the time to look at all of them carefully...

Using "triangular" number denominators, he contends that

pi - 2 = 1/1 + 1/3 - 1/6 - 1/10 + 1/15 + 1/21 - ...

Notice the two positive terms followed by two negatives, etc., something rather unusual...and "triangular" numbers are those which are generated by n(n+1) /2...

His proof goes soemthing like this...let A equal the sum of the odd terms and B be the sum of the even terms, that is:

A = 1/1 - 1/6 + 1/15 - ...
B = 1/3 - 1/10 + 1/21 - ...

Now A = 2/(1 * 2) - 2/(3 * 4) + 2/(5 * 6) - 2/(7 * 8) + ...

which is also equal to

A = (2/1 - 2/2) - (2/3 - 2/4) + (2/5 - 2/6) - (2/7 - 2/8) + ...

Now if we unite the terms with even denominators, he gets

- 1/1 + 1/2 - 1/3 + 1/4 - 1/5 + 1/7 - ...

and that is equal to (- log 2) according to the well-known expansion

log (1 + x) = x - x^2/2 + x^3/3 - x^4/4 + x^5/5 - ....

The terms with the odd denominators is

2/1 - 2/3 + 2/5 - 2/7 + 2/9 - 2/11 + ...

that is equal to pi/2, according to a well-known expansion for pi...

therefore A = pi/2 - log 2.

Okay, now with

B = 2/(2 * 3) - 2/(4 * 5) + 2/(6 * 7) - 2/(8 * 9) + ...

that is equal to

B = (2/2 - 2/3) - (2/4 - 2/5) + (2/6 - 2/7) - (2/8 - 2/9) + ...

If we unite the terms with even denominators we get

1/1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 + 1/7 - ...

that he is equal to log 2.

The terms with odd denominators are equal to

- 2/3 + 2/5 - 2/7 + 2/9 - 2/11 + 2/13 - 2/15 + ...

that he is equal to pi/2 - 2

So therefore B = pi/2 - 2 + log .

Now let's put it together:

If we unite the two parts A and Bm we get

A + B = pi/2 - log 2 + pi/2 - 2 + log 2 = pi - 2.

Pretty neat, huh?

And here is even more...PiHobbes.txt...

Thanks to Candido Otero for bringing it to my attention.

More from Mr. Otero: Here are attachments ARCTANGENT.PDF ARCSINE.PDF and PI_HOBBES.PDF,
where are the new formulas that calculate the functions, Arctangent and Arcsine,
and the formula that is in the file PI_HOBBES.PDF to calculat PI divided by
any real number x.

The other files, PIHOBBES2.PDF PIHOBBES3.PDF and HOBBES.PDF, are special cases that
apply to the formula that is in the file, PI_HOBBES.PDF.

The Phenomenon of CoPrimes (thanks to Eileen's son)

Imagine you have two big piles of stamps.  One pile is made up of stamps worth 3¢ each, while the other pile consists entirely of 5¢ stamps.  
This year, postage is 8¢ — but the postal service has decreed that postage will increase by 1¢ each year for the rest of time (thanks, Obama),
and you must pay the exact price.  So next year, postage will be 9¢, the year after that 10¢, then 11¢, and so on.  Since you only have 3¢
and 5¢ stamps, what years won’t you be able to send letters because you can’t hit the exact cost of postage?

The answer is never.  You will never be unable to send a letter, because any positive integer greater than 7 can be formed with some combination of 3s and 5s:
8 = 3 + 5
9 = 3 × 3
10 = 5 × 2
11 = 5 + 3 × 2
1,236 = 5 × 246 + 3 × 2

This is a ridiculously cool phenomena associated with coprime numbers — numbers whose only common divisor is one.  Two coprime numbers a and b 
can combine to form any positive integer whose value is greater than ab – (a+b).  Combinations of 3 and 5 can make any number greater than 7; 2 and 7
can make any number greater than 5; 73 and 182 can make any number greater than 13,031; and so on...

The Birthday Problem (thanks to Eileen's son)


Cool little bit of probability that would be fun to pull out at parties, if you go to the right kind of parties — or wrong kind of parties, probably (get it? probably? eh? ehhh?):
In a room full of fifty people, what is the chance that two people have the same birthday?
Think about it for a second.  There are 365 days in the year (this is an idealized mathematical world, so leap years don’t exist), and only fifty people in the room —
not nearly enough to ensure that every day is covered, right?

Well, right, but that’s not what the question is asking.  We want to know whether two people share a birthday, not whether two people are born on a certain day.
Assuming a uniform distribution of birthdays over the year (which isn’t really totally true, but whatever), let’s find the chance that two people in the room don’t share
a birthday.  To do that, we start with the second person in the room.  The chance that the second person shares a birthday with the first person in the room is one out of 365 —
but the chance they don’t share a birthday is 364/365.  The chance the third person does not share a birthday with the first or the second person is 363/365.

The fourth person has a 362/365 chance of not sharing a birthday, and so on, until you’ve gotten to the fiftieth person, who has a 316/365 chance of being born on a unique day.
Since everyone’s birthdate is independent of one another’s (for simplicity we’re assuming no weird parental pregnancy pacts between the parents of people in the room), to find the total probability that two people don’t share a birthday we have to multiply all these individual probabilities together.  Imagine flipping a coin, which is the archetypal independent event — the chance of getting heads is 1/2, but the chance of getting two heads in a row is 1/2 x 1/2 = 1/4, and so on.  That means we have:

Or, using the factorial operator and rearranging:

Since probabilities always have to add up to 100%, the probability that at least two people share a birthday is one minus this result — which works out to 97% (!).  That means that, in a random assortment of just fifty people, having two people share a birthday is almost a certainty.

In fact, at only 23 people in a room, there’s over a 50% chance that two of them will share birthdays.  Math!