Click to read the comments on this entry...
Bullseye optical illusion
Aug 5, 2010Multiple choice question
Jul 31, 2010
Multiple Choice:
If you choose an answer to this question at random what is the chance you will be correct?
Click to read the comments on this entry...If you choose an answer to this question at random what is the chance you will be correct?
A) 25%(discussion at the Math Subreddit)
B) 50%
C) 60%
D) 25%
The Bailey-Borwein-Plouffe formula
Jul 22, 2010
This is an excellent story about the so-called Bailey-Borwein-Plouffe algorithm and formula. The BBP formula is an expression for calculating pi discovered by Simon Plouffe in 1995:
This is a remarkable formula because it is a digit-extraction algorithm for pi in base 16. Simon Plouffe describes his discovery of the formula and the consequences he suffered when sharing the credit with others. Read the full story here or the highlights here (the original story is very lengthy):
Click to read the comments on this entry...
This is a remarkable formula because it is a digit-extraction algorithm for pi in base 16. Simon Plouffe describes his discovery of the formula and the consequences he suffered when sharing the credit with others. Read the full story here or the highlights here (the original story is very lengthy):
"The story began many years ago in 1974 when I wanted to find a formula for the n'th digit of Pi... since the computation of Pi looks more complicated than the number e... I studied a way to compute that number instead.
...During my stay at Bordeaux University in 1992-1993 I perfected that program I had that could interface Pari-Gp and Maple. That little Unix script had an enormous advantage of flexibility because I could set up a series of real numbers to test among 1 unknown. At that time I was beginning to find new results, the programs were able to find identities.
That program was the one that found the formula for Pi in hexadecimal (or binary)...
This is where I made the biggest mistake in my life : To accept the collaboration of Peter Borwein and David H. Bailey as co-founders of that algorithm and formula when they have found nothing at all. David Bailey was not even close to me when I found the formula. He was added to the group 2 months after the discovery.
I was naively thinking that I could negotiate a job as professor at Simon Fraser University, which failed. I am very poor at negotiations. I remember that day when the Globe & Mail newspaper article went out in October 1995. I was at Jon Borwein's house and he had a copy of the newspaper in hand. This is where I asked him to become a professor at SFU. He simply replied right away < don't even think about it >. I thought, this is the best chance I will ever have to become a professor there, since it failed, I decided that I had to leave that place.
I was very frustrated at that time, in late 1995 after the discovery. I realized that many small details where terribly wrong. They were getting a lot of credit for the discovery and I had the impression of not getting anything in return. My strategy failed... Later that year, I was invited to a ceremony in Vancouver for the CUFA (faculty of the year Award). This is a prize with plaque and mention that those 2 brothers received for the discovery of the formula. They simply mentioned my name at the ceremony and I received nothing at all...
Then in 1996, I realized that if I get up at night to hate them it is a very bad sign, it means that I have to leave that place (Simon Fraser university). I was convinced I had no future at all with those 2 guys around. I was making serious plans to leave.
...About David H. Bailey. He came after the discovery of the formula and my small basic program , I had also a Fortran version. This is where Peter Borwein suggested to add him as a collaborator to the discovery since he contributed to it (as he said), this is my second big mistake. Of course he accepted to co-write the article, who wouldn't ?! David H. Bailey (and Ferguson) are the authors of the PSLQ program. That program is theversion of the Pari-Gp program. I used it a little it is true, but what made the discovery was Pari-Gp and Maple interface program I had. So actually, that person has nothing to do with the discovery of that algorithm and very little to do with the finding of the formula. The mistake was mine. Saying that Bailey found the formula is like saying that the formula was found by the Maple and Basic program. I tried very hard to correct the situation avoiding the subject of the actual discovery of the algorithm and the formula, I made an article in 1996 for the base 10. I thought naively again that this would re-establish the situation, it did not. I almost accepted to do a film at one point in 1999 when a certain guy from England that wanted to make a movie on Pi and the discovery of the formula. he asked me if I would accept to talk about my with the Borweins. I did not wanted to go in that direction, I should had. There was that book of Jean-Paul Delahaye (le fascinant nombre pi) that mentioned the Plouffe algorithm and formula because I told him part of the story. In some way I was afraid of revealing that enormous story. Why was I so naive? I had a previous collaboration with Neil Sloane and the Encyclopedia of Integer Sequences and the web site, this was really a big success and Neil is the person I respect the most in mathematics so this is why I thought (wrongly ) that my collaboration with the Borweins had to go well, a big mistake. Why do I write this? To tell the truth and also the arrogance of those people makes me sick. Will I gain something from this? I don't care, I have nothing to loose. Simon Plouffe Montréal, le 22 juin 2003."
Sine Graph
Jul 16, 2010
See if the above makes sense. It illustrates that cos(x) is indeed the derivative of sin(x).
Four time lottery winner
Jul 13, 2010
This is interesting... Joan Ginther from Texas, has won multi-million dollar lottery jackpots four times so far!
That's close to $21 million in winnings for Joan! Mathematicians (namely Sandy Norman and Eduardo Duenez) say the chances of winning four multi-million dollar jackpots are as slim as 1 in 18 septillion -- that's
- In 1993 she split $11 million.
- In 2006 she won $2 million.
- In 2008 she won $3 million.
- Last month, she won $10 million.
1 in 18,000,000,000,000,000,000,000,000!
The interesting thing is that Joan Ginther herself was a professor (she earned her doctorate from Stanford University in 1976) and spent many years teaching math.
This is neat. Asur and Huberman (HP Labs California) have analyzed the rate of tweeting versus opening weekend box office revenue. As expected, they found a strong correlation between the amount of tweets concerning a forthcoming film, and its opening weekend box office return (lots of people talking about it, means lots of people want to see it, right?). MrScienceShow says:
"After examining the rate of chatter from almost 3 million movie tweets, the researchers constructed a linear regression model for predicting box-office revenues of movies in advance of their release. These results outperformed the Hollywood Stock Exchange, a market in which people can buy and sell virtual shares in actors, directors and individual movies and produces unusually accurate predictions of film popularity."The results can be found in their paper on the arxiv.
Hidden Matlab Fun (Easter Eggs)
Jun 17, 2010
Have some fun with Matlab today. Here are some easter eggs for the program (it may depend on your version, but all of these worked on mine).
Open the Matlab command window and type one of the following to get a nice surprise:
>> image
This one gives an image of a boy.
>> spy
This one gives the classic spy vs spy picture.
>> life
Simulation of Conway's Game of Life.
>> why
Alternatively type why(n), for some number n.
Repeat.
>> penny
American penny.
>> toilet
Simulation of a working toilet, haha.
Also try "shower".
Click to read the comments on this entry...Open the Matlab command window and type one of the following to get a nice surprise:
>> image
This one gives an image of a boy.
>> spy
This one gives the classic spy vs spy picture.
>> life
Simulation of Conway's Game of Life.
>> why
Alternatively type why(n), for some number n.
Repeat.
>> penny
American penny.
>> toilet
Simulation of a working toilet, haha.
Also try "shower".
Sphericons
Jun 7, 2010
This is neat.... a sphericon is a 3-dimensional geometric solid with a single continuous face and two edges (sort of like a Möbius strip). The sphericon rolls along a wobbley path but ends up traveling in a straight line!
Over at Fold Play, someone turned a pie into a sphericon - watch as it rolls in a straight line down the slope. Pretty neat.
They also have templates to make your own sphericon pie.
Click to read the comments on this entry...Over at Fold Play, someone turned a pie into a sphericon - watch as it rolls in a straight line down the slope. Pretty neat.
They also have templates to make your own sphericon pie.
From the ScienceDaily:
Michael Doebeli, Iaroslav Ispolatov. Complexity and Diversity. Science, 2010: 328 (5977): 494-497 DOI: 10.1126/science.1187468
Keep up the good work mathematicians! :D
Click to read the comments on this entry..."UBC researchers have proffered a new mathematical model that seeks to unravel a key evolutionary riddle--namely what factors underlie the generation of biological diversity both within and between species."Their research can be found in Science:
Michael Doebeli, Iaroslav Ispolatov. Complexity and Diversity. Science, 2010: 328 (5977): 494-497 DOI: 10.1126/science.1187468
Keep up the good work mathematicians! :D
Freshman's dream and sophomore's dream
Apr 23, 2010
For those who haven't heard of this yet, the freshman's dream is given to the (common) error:
where n is usually a positive integer greater than 1 (can be real too). You'd be surprised how many university students make this mistake! Simplying looking at n=2 shows why it doesn't work in general: (x + y)2 = x2 + 2xy + y2. However, there is a theorem referred to as the "Freshman's Dream" which says if p is a prime number, and x,y are members of a commutative ring of characteristic p, then (x + y)p = xp + yp.
The sophomore's dream is used for the following identity:
This formula was discovered in 1697 by J. Bernoulli. The sophomore's dream seems too good to be true (like the freshman's dream), but is in fact true!
Click to read the comments on this entry...(x + y)n = xn + yn,
where n is usually a positive integer greater than 1 (can be real too). You'd be surprised how many university students make this mistake! Simplying looking at n=2 shows why it doesn't work in general: (x + y)2 = x2 + 2xy + y2. However, there is a theorem referred to as the "Freshman's Dream" which says if p is a prime number, and x,y are members of a commutative ring of characteristic p, then (x + y)p = xp + yp.
The sophomore's dream is used for the following identity:
This formula was discovered in 1697 by J. Bernoulli. The sophomore's dream seems too good to be true (like the freshman's dream), but is in fact true!
Shoe Lacing Math
Apr 20, 2010
How many ways are there to tie your shoes?
Depends on the shoes, but Ian has done some calculations (under certain assumptions) to show there are over 2 trillion ways!
Mathematician, Burkard Polster, published an article in the journal Nature (December 2002) about the mathematics of shoe lacing. His calculation for the number of 'real-world' lacing methods (of a shoe with 12 eyelets) is 43,200. He also has a shoe lace book that is published by the AMS.
Bike with square wheels
Apr 13, 2010
I'm back to my regular update schedule. I was just visiting Macalester College and saw that they had the tricycle pictured below! I have seen pics on the internet of it but didn't know where it existed.
The original was created in 1997 (and replaced in 2004) by Professor Wagon (not Professor Bike) and rolls smoothly over a road made out of inverted "catenaries". See a video here for a demonstration.
Click to read the comments on this entry...
The original was created in 1997 (and replaced in 2004) by Professor Wagon (not Professor Bike) and rolls smoothly over a road made out of inverted "catenaries". See a video here for a demonstration.
Making change
Apr 3, 2010
Consider the following hypothetical situation:
You are a cashier and you have to give the customer $0.40 in change, however, you have no nickels left in your till. The goal is to minimize the number of coins that the customer receives. How do you do it?
Well this actually happened, and the process the cashier went through was to:
This produces 7 coins! The cashier realized this was not good, thought for a moment, then grabbed 4 dimes, which is the optimal solution in this case.
- first take out a quarter (the highest denomination coin below $0.40),
- then take out a dime (the next highest denomination below $0.15),
- then finally, take out 5 pennies (since there is no nickels).
The "Change-making problem" is a well known problem you may have encountered if you have studied optimization. The problem is to see how one can give change with the least number of coins of given denominations.
The interesting thing is that for currency in North America, the greedy algorithm always produces an optimal solution (i.e. picking the largest denomination of coin which is not greater than the remaining amount). As we saw in the above example, if nickels were not allowable coins, the greedy algorithm would no longer produce an optimal solution for our currency denominations.
Check out the wikipedia links above for more info about this problem :P
The first one is how to cheer up your friend...
Just don't tell him he should have one ovary :P
This actually appeared in a textbook...
Click to read the comments on this entry...
This actually appeared in a textbook...
Deck Shuffling
Mar 31, 2010
Cool fact: If you thoroughly shuffle an ordinary deck of 52 playing cards, chances are that the resulting arrangement of cards has never existed before (via Reddit and Reddit).
Quoting from the math subreddit:
"Proof -- easy numeric comparison. There are 52! possible orderings of a deck, and I'm assuming all are equally likely after your shuffling. Let's wildly overestimate and assume that every second since the universe was created, a million decks of cards were shuffled and someone looked through them. Thus fewer than 10^24 orderings have ever been seen.Discussion here.
But at an incredibly crude estimate, 52! is at least 10^42 * 10!; let's underestimate that again wildly by 10^42. That means that chances of your ordering ever having come up previously are at most 1 in 10^18.
(Note, by the birthday paradox, the chances that there have been two identical orderings observed by two people in history are quite a bit higher -- perhaps even feasibly likely; I haven't calculated it. But we're looking here at the probability that a given ordering matches one of the ones previously seen.)"
Memorizing Pi - World Records and Techniques
Mar 15, 2010
How many digits of pi do you have memorized?

But seriously... is it 3? 5? 10? more than 30? If it's more than 30 pat yourself on the back because that's a great accomplishment! If it's only a few, then no worries. Below we will teach you some techniques that can be used to conquer the digits of pi.
Computations of Pi
Some basic information and a brief time line on computations of the digits of pi:
Who memorizes pi?
This is just a joke. It does bear a tiny bit of truth but the two sets of people aren't mutually exclusive. I am both a scientist and a science fan!
Digits Memorized vs. Year (Graph)
Record Holders*: David Fiore
April 1st, 1979:
Record Holders*: Creighton Carvello (1944-2008)
June 27th, 1980:
Record Holders*: Rajan Mahadevan
July 5, 1981:
Record Holders*: Hideaki Tomoyori
March 10th, 1987:
Record Holders*: Chao Lu
November 20th, 2005:
Unofficial: Andriy Slyusarchuk
June 17th, 2009:
Why memorize pi? To beat Grace!
May 12th, 2008:
How to memorize pi? Piems!
A piem is a (pi) poem where the length of each word represents a digit of pi
For example, the following piem encodes the string: 3. 141592 65358 9793 23846
Pie = 3;
I = 1; wish = 4; I = 1; could = 5; determine = 9; pi = 2;
Eureka=6; cried=5; the=3; great=5; inventor=8;
and so on. Thus, each word represents a digit of pi.
My favourite piems!
There's over a bazillion piems and variations (lots and lots). The best ones are:
May I have a large container of coffee right now?
3.141592653
Hey, I need a large motorboat to rescue women and girls.
3.1415926535
God! I need a drink,
Alcoholic of course,
After all those lectures
Involving radical equations.
3.1415 926 5358 979
Long Piems
Technique: Grouping Digits
Start by memorizing (3.14159) for a minute... then add the next group (26535) and practice for two minutes. Then add the third group and practice until you are comfortable (REPEAT!!)
Classic Memory Techniques - The Major System
Classic Memory Techniques - Link System
You are standing at the biggest door you have every seen.
You knock at the door and this Raggedy Ann doll answers.
Out of nowhere, she smacks you with a pan she is holding!
Coordinate Method
Click to read the comments on this entry...
But seriously... is it 3? 5? 10? more than 30? If it's more than 30 pat yourself on the back because that's a great accomplishment! If it's only a few, then no worries. Below we will teach you some techniques that can be used to conquer the digits of pi.
Computations of Pi
Some basic information and a brief time line on computations of the digits of pi:
- 1540 - 1610: 35 digits determined
- done by German mathematician Ludolph van Ceulen
- used a geometric method (just like Archimedes did)
- proud of his calculation that took a great part of his life
- he had the digits engraved on his tombstone
- 1949: 2, 037 digits computed (John von Neumann et al.)
- 1973: Over one million digits computed
- 1989: One billion digits computed (Chudnovsky brothers)
- 2010: 2.7 trillion digits computed (F. Bellard)
- In the near future: Almost all of them computed?
Who memorizes pi?
This is just a joke. It does bear a tiny bit of truth but the two sets of people aren't mutually exclusive. I am both a scientist and a science fan!
Digits Memorized vs. Year (Graph)
Record Holders*: David Fiore
April 1st, 1979:
- David Fiore wrote down 10,625 decimal places of pi
- He was 18 years old at the time
- He is known as the first person to ever break 10,000 decimal places
- It took him three hours and five minutes
Record Holders*: Creighton Carvello (1944-2008)
June 27th, 1980:
- Creighton Carvello recited 20,013 decimal places of pi
- 2003: he recalled 3,500 facts about every FA Cup Final since 1872 (names of referees, goal scorers, teams, crowd attendances, scores, venues...)
- Memorized the exact sequence of 10,000 words from Ernest Hemingway's The Old Man and the Sea
- Recited 17 random digits after seeing them for 2 seconds
Record Holders*: Rajan Mahadevan
July 5, 1981:
- Rajan Mahadevan recited 31,811 digits of pi
- He discovered his exceptional ability to memorize numbers at the age of 4 during a party hosted by his family
- During the party, Rajan wandered to a parking lot and committed the license plate numbers of every guest's car for recitation later
- A quote: "I am not good at remembering words - words confuse my system of memorizing. Numbers, I have no problems at all. I put away huge numbers in something similar to a computer file and I can recall them even after decades."
Record Holders*: Hideaki Tomoyori
March 10th, 1987:
- Hideaki Tomoyori recited 40,000 decimal places of pi
- Took him 17 hours 21 minutes (including breaks totaling 4 hours 15 minutes) to recite
- Took him 10 years to memorize 40,000 decimal places
Record Holders*: Chao Lu
November 20th, 2005:
- Chao Lu recited 67,890 decimal places of pi
- Took him 24 hours 4 minutes to recite (with no breaks)
- Took him 1 year to memorize 100,000 digits (he made a mistake at the 67,891th digit when going for the record)
- He is the current (official) record holder
- In 2006, Akira Haraguchi, a retired Japanese engineer, claimed to have recited 100,000 decimal places. This, however, has yet to be verified by Guinness World Records.
Unofficial: Andriy Slyusarchuk
June 17th, 2009:
- A. Slyusarchuk claims to have 30 million digits memorized
- The digits are printed in 20 volumes of text
- He is a neurosurgeon, medical doctor and professor
- He was able to recite randomly selected sequences from within the first 30 million places of pi
- Reciting 30 million digits of pi at one digit a second would take 347 days (nonstop)
- No officially documented attempt to debunk his claims has been successful as of yet
Why memorize pi? To beat Grace!
May 12th, 2008:
- Grace Hare recited 31 digits of pi
- It took her 18 seconds
- She is 3 years old and the youngest record holder
How to memorize pi? Piems!
A piem is a (pi) poem where the length of each word represents a digit of pi
For example, the following piem encodes the string: 3. 141592 65358 9793 23846
PieNotice that:
I wish I could determine pi
Eureka! cried the great inventor.
Christmas pudding, Christmas pie
Is the problem's very center.
Pie = 3;
I = 1; wish = 4; I = 1; could = 5; determine = 9; pi = 2;
Eureka=6; cried=5; the=3; great=5; inventor=8;
and so on. Thus, each word represents a digit of pi.
My favourite piems!
There's over a bazillion piems and variations (lots and lots). The best ones are:
May I have a large container of coffee right now?
3.141592653
Hey, I need a large motorboat to rescue women and girls.
3.1415926535
God! I need a drink,
Alcoholic of course,
After all those lectures
Involving radical equations.
3.1415 926 5358 979
Long Piems
- The short story Cadaeic Cadenza encodes 3835 digits
- It was written in 1996 by Mike Keith
- Words of length 10 encode the digit 0
- Words of length 11 (or 12) encode the two consecutive digits 1,1 (or 1,2)
- 2010: In his book Not A Wake, Keith extends to 10,000 digits of pi
Technique: Grouping Digits
- Split pi into small groups of digits (like 4 digits or 5, 6, 7, whatever you are comfortable with)
- Focus on memorizing the first small group
- Some people find singing it helps
- When comfortable with the first group, move on to the next
- Cons: If you lose your spot, you may have to start over.
Start by memorizing (3.14159) for a minute... then add the next group (26535) and practice for two minutes. Then add the third group and practice until you are comfortable (REPEAT!!)
Classic Memory Techniques - The Major System
- Major System: Convert numbers into sounds.
- Sounds without numbers are used as 'fillers'
- Form words from the sounds
- In practice, use 100 'peg words': rat is 41; bar is 94
Classic Memory Techniques - Link System
- Start by converting each digit of pi to its corresponding phonetic sound
- Group sounds together to create a list of words
- Words created should be actions or objects
- Alternatively, use your 'fixed' peg words for the number
- Use the Link System: Link words together into a long chain by using a sequence of events, a story, or a journey. The CRAZIER the story the BETTER!!
You are standing at the biggest door you have every seen.
You knock at the door and this Raggedy Ann doll answers.
Out of nowhere, she smacks you with a pan she is holding!
Coordinate Method
- Pros: Can recite starting at any decimal spot (if you lose your spot, you don't have to start over)
- First 10 decimal places (1415926535) associated with 0
- Use the Major System to encode as: turtle-pinochle-mall and link it to 0 (saw)
- Example: Picture yourself using a saw to cut open a turtle who is playing pinochle at the mall
- Next 10 digits (8979323846) would be linked to 1 in the same manner
- Next 10 digits linked to 2
- Repeat.
Topological Poker
Feb 28, 2010
This is interesting. RandomPairing over at the math subreddit listed some rules for a game called Topological Poker (aka Dicks and Hoes), which I assume he created himself. [I did a google search for "Topological Poker" and only got the subreddit, and then did a search for "Dicks and Hoes" and got a bunch of dirty sites but no poker ones.]
RandomPairing explains...
Go to the reddit link to find out the rules of play."There are four ranks of cards, instead of the usual 13. Instead of 2 through Ace, cards are ranked according to the topological classification of their symbol ... 2,3,5,7,J,K all have the same property that they can be deformed into a dot or line. These are the lowest ranked cards. In fact, by themselves they're worth dick--hence the name for these cards.
4,6,9,A are topologically the same as a 'o', and hence are called holes, ohs, or--if you can get away with it--hoes. These are the second ranked cards in ascending order.
10 is in a class by itself because it's a combination of a dick and a hoe, so it's called a Split, and is third ranked.
8,Q are topologically the same because of the way the Q is made on cards, with the slash going all the way across and creating two holes. This is the highest rank--the double hoe, or just double."
Fibonacci Spiral
Feb 22, 2010Why asians rock at math
Jan 28, 2010Dudeney numbers
Jan 16, 2010
This is pretty neat. A Dudeney number is a positive integer that is a perfect cube such that the sum of its decimal digits is equal to the cube root of the number. For example:
How many such numbers do you think exist?
Well, it turns out there is exactly 6. They are: 1, 512, 4913, 5832, 17576 and 19683.
Click to read the comments on this entry...5832 = 18 x 18 x 18... and 5 + 8 + 3 + 2 = 18.
How many such numbers do you think exist?
Well, it turns out there is exactly 6. They are: 1, 512, 4913, 5832, 17576 and 19683.
Happy birthday Hawking, happy death day Galileo
Jan 8, 2010
Just a note that today is Stephen Hawking's birthday and also Galileo's death day.
Galileo Galilei
Born: Feb 15th, 1564 (Pisa)
Died: Jan 8th, 1642 (Arcetri)
Stephen William Hawking
Born: Jan 8th, 1942 (Oxford)
Died: Should have died five decades ago.
Hawking's birth occurred on the 300th anniversary of Galileo's death!
Hawking says, "Galileo, perhaps more than any other single person, was responsible for the birth of modern science."
Click to read the comments on this entry...Galileo Galilei
Born: Feb 15th, 1564 (Pisa)
Died: Jan 8th, 1642 (Arcetri)
Stephen William Hawking
Born: Jan 8th, 1942 (Oxford)
Died: Should have died five decades ago.
Hawking's birth occurred on the 300th anniversary of Galileo's death!
Hawking says, "Galileo, perhaps more than any other single person, was responsible for the birth of modern science."
Perfect Pizza Slicing
Dec 16, 2009
This has honestly been all over the internet. EVERY single blog/website I read had this -- boing boing, slashdot, digg, digg again, mixx, reddit (posted 4 times actually), among with a billion other sites.
And what kind of post would this be if I didn't include a picture. Blue pizza yum:
Click to read the comments on this entry...PIZZA MATH!!!
AHHHH. Enough with the stupid pizza slicing!!. Anyways, if you seriously haven't seen this yet then it's at the link below:PIZZA MATH!!!
PIZZA MATH!!!
New Scientist Link"Suppose the harried waiter cuts the pizza off-centre, but with all the edge-to-edge cuts crossing at a single point, and with the same angle between adjacent cuts. The off-centre cuts mean the slices will not all be the same size, so if two people take turns to take neighbouring slices, will they get equal shares by the time they have gone right round the pizza - and if not, who will get more?Of course you could estimate the area of each slice, tot them all up and work out each person's total from that. But these guys are mathematicians, and so that wouldn't quite do. They wanted to be able to distil the problem down to a few general, provable rules that avoid exact calculations, and that work every time for any circular pizza."
And what kind of post would this be if I didn't include a picture. Blue pizza yum:
Pi crop circles
Dec 13, 2009
You may have seen this already since it dates back to June 2008... But there was reports of a crop circle that represented the first 10 digits of pi (including the decimal point). Pictures below.
Believers in aliens argue it was made by mathematically minded aliens... But of course, there is a super high probability it wasn't.
One thing is that it is in base 10. If aliens made it, why would they use base 10 unless they have been studying our race for a long time. In which case, why would they leave silly circles in random fields? Also, the smart aliens live in England, since these kinds of mathematical crop circles have appeared there a few times.
Click to read the comments on this entry...
One thing is that it is in base 10. If aliens made it, why would they use base 10 unless they have been studying our race for a long time. In which case, why would they leave silly circles in random fields? Also, the smart aliens live in England, since these kinds of mathematical crop circles have appeared there a few times.
Pinocchio paradox
Dec 9, 2009Pi explaination
Dec 6, 2009
This is neat hahaha... No idea where it came from. It says gifbin.com on the image though :D
Do the math - weightloss
Dec 3, 2009
Want to lose 10 pounds within two months? All you have to do is the math!!
My simple 5 step program is flawless I tell ya, FLAWLESS!
My simple 5 step program is flawless I tell ya, FLAWLESS!
-
Multiply the number of pounds you want to lose by the number of calories in a pound of fat. 3500 calories in a pound of fat
So, if you want to lose 10 pounds, it comes out to 35,000 calories. - Divide this total by the days you will be on the diet. This will give the deficit you need to create in your diet. In our 2 month program example, we will do:
Deficit = 35,000 ÷ 60 = 583.
- Calculate your Basal Metabolic Rate using the formula below:
For Men:
BMR = 66 + (13.7 X wt in kg) + (5 X ht in cm) - (6.8 X age in years)
For Women:
BMR = 655 + (9.6 X wt in kg) + (1.8 X ht in cm) - (4.7 X age in years)
Conversion:
1 inch = 2.54 cm.
1 kilogram = 2.2 lbs.
Legend:
wt = weight.
ht = height.
For example, my BMR is equal to 1730.
- Calculate your total daily energy expenditure (TDEE) by multiplying your BMR by your activity multiplier from the chart below:
Activity Multiplier
Sedentary = BMR X 1.2 (little or no exercise, desk job)
Lightly active = BMR X 1.375 (light exercise 1-3 days/wk)
Mod. active = BMR X 1.55 (moderate exercise 3-5 days/wk)
Very active = BMR X 1.725 (hard exercise 6-7 days/wk)
Extr. active = BMR X 1.9 (hard daily exercise & physical job or 2X day training, e.g. marathon)
So, TDEE = BMR x (number from above).
For me, TDEE = 1730 x 1.375 = 2378.
- Then use the formula:Allowed Calories Per Day = TDEE - Deficit
For me, Allowed Calories Per Day = 2378 - 583 = 1795. Thus, for me to lose 10lbs over the course of 2 months, I would need to consume roughly 1800 calories per day (and NO MORE). I tried it out and it worked for me (the math does not lie!!).
Remember to eat low calories foods whenever you can. Substitute light beer for regular beer, if you HAVE to drink beer. Eat "sugar free" jello, as its only 5 calories per cup. Celery is practically 0 calories. If you have to drink soft drinks, choose diet, as it's only 2-3 calories per can. Take just milk in your coffee, no sugar (or drink it black as it's then essentially 0 calories). Drink lots of water, LOTS. You don't have to lay off the fast food, so if you're eating out, try to look at the number of calories in some of the meals and compare. Choose the lowest. You can go to burger king and eat an entire fast food UNHEALTHY meal for a total cost of 500 calories in your diet. That's pretty good, because then you'll still have 1000+ calories to consume for the day. (Though, I don't recommend eating fast food all the time arrg).
A Friendly Chat About Whether 0.999... = 1
Dec 1, 2009
Via reddit I stumbled upon this site which talks about something called hyperreal numbers and claims that in this theory, the equation 0.999... = 1 is false.
For those who follow the internet, the question of whether 0.999...=1 has come up a quadbrazillion times on practically every math related forum and even the non-math ones. And every single time it turns into this huge argument with
non-mathematicans vs frustrated-mathematicians
and neither of them winning. Some mathie forums even have strict rules banning users against posting topics that deal with 0.999....=1.But the site mentioned above actually gives some decent points to support why 0.999...=1 isn't necessarily true., but you have to change some of the concepts that we take for granted.
In my opinion, 0.999... and 1 are as equal as can be.
Are the reals countable?
Nov 29, 2009
John Gabriel (a non-mathematician) claims in his blog that the real numbers are countable. The author tries to enumerate the real numbers in the interval [0, 1) by writing out all those whose decimal representation has one digit after the dot (0.1, 0.2, ..., 0.9), followed by those with two digits after the dot, then those with three digits, and so on.
He goes on further to say that this establishes a procedure for writing
down an ordered sequence of numbers in which every real number of the
source interval will appear eventually (although, it should be clear
that any number with an infinite decimal representation will never
occur in his sequence of numbers).
This argument shows that the subset of real numbers between 0 and 1 that have a finite decimal representation is countable. Although, it fails to work for the interval [0, 1).
It's rather entertaining to watch fellow mathematicians humor him. Either way, things like this happen all the time :-D.
Click to read the comments on this entry...This argument shows that the subset of real numbers between 0 and 1 that have a finite decimal representation is countable. Although, it fails to work for the interval [0, 1).
It's rather entertaining to watch fellow mathematicians humor him. Either way, things like this happen all the time :-D.
On a curious property of 3435
Nov 23, 2009
This is a neat little paper regarding the number 3435. Basically it has the following property:
More interesting is that the only other number with this 'property' is the number 1. The property is what he calls a "Munchausen Number". The preprint was written by Daan van Berkel and can be found on the arxiv.
Click to read the comments on this entry...3435 = 3^3 + 4^4 + 3^3 + 5^5
More interesting is that the only other number with this 'property' is the number 1. The property is what he calls a "Munchausen Number". The preprint was written by Daan van Berkel and can be found on the arxiv.
Guess 2/3 of the average
Nov 11, 2009
The game is...
- you compete against some other people
- each of you guess one number from [0, 100]
- compute 2/3rd's of the average of the guessed numbers
- the winner is whoever is closest
(0.66666) x ([7 + 28 + 53 + 77] / 4) = 27.5
Therefore, whoever guessed 28 wins!!
So........... what number should you pick?
Of course the answer depends on your thought process. Let's take a look to see what numbers people picked:

So what number would you pick?
One should note that guessing any number that lies above 66.66 can NEVER be equal to 2/3rd's the average. Why is this true? Take the example that everyone picks 100. Then the average is 100. So 2/3rd's of this is 66.66. This is the highest that "2/3rd's the average" can be.
Thus, any rational player, would pick a number between [0, 66.67].
Now, any rational player would realize that everyone else is going to pick a number in [0, 66.67]. And thus the highest that 2/3rd's the average can be using numbers in [0, 66.67] is 44.444. So the rational person will not pick a number above 44.444, since there is no way it can be 2/3rd's the average.
Thus, any rational player, would pick a number between [0, 44.44].
Now you can see you can repeat this process and eventually get yourself down to [0, epsilon]. Thus, if everyone was rational, they would all pick the number 0, and hence, it would be a tie game.
However, strangely enough, not everyone can work the argument all the way down to 0. Hence, you have people picking 22 and 33 (see the spikes in the graph). The amazing thing is that people pick numbers like 100! There is no way 100 can win, so clearly these people don't understand the game at all, or never even thought about their answer. There is also a spike at 0, most likely by mathematicians and people from similar disciplines who are familiar with the problem. However, if they were really smart, they'd know that they can't win picking 0, since it's highly likely that not everyone will pick 0.
Click to read the comments on this entry...Of course the answer depends on your thought process. Let's take a look to see what numbers people picked:

So what number would you pick?
One should note that guessing any number that lies above 66.66 can NEVER be equal to 2/3rd's the average. Why is this true? Take the example that everyone picks 100. Then the average is 100. So 2/3rd's of this is 66.66. This is the highest that "2/3rd's the average" can be.
Thus, any rational player, would pick a number between [0, 66.67].
Now, any rational player would realize that everyone else is going to pick a number in [0, 66.67]. And thus the highest that 2/3rd's the average can be using numbers in [0, 66.67] is 44.444. So the rational person will not pick a number above 44.444, since there is no way it can be 2/3rd's the average.
Thus, any rational player, would pick a number between [0, 44.44].
Now you can see you can repeat this process and eventually get yourself down to [0, epsilon]. Thus, if everyone was rational, they would all pick the number 0, and hence, it would be a tie game.
However, strangely enough, not everyone can work the argument all the way down to 0. Hence, you have people picking 22 and 33 (see the spikes in the graph). The amazing thing is that people pick numbers like 100! There is no way 100 can win, so clearly these people don't understand the game at all, or never even thought about their answer. There is also a spike at 0, most likely by mathematicians and people from similar disciplines who are familiar with the problem. However, if they were really smart, they'd know that they can't win picking 0, since it's highly likely that not everyone will pick 0.
How to: organize your digital library of papers
Oct 29, 2009
There are a few ways you can organize the digital papers you download.
1. One such way is using Zotero. Basically, Zotero [zoh-TAIR-oh] is a free, easy-to-use Firefox extension to help you collect, manage, and cite your research sources.
1. One such way is using Zotero. Basically, Zotero [zoh-TAIR-oh] is a free, easy-to-use Firefox extension to help you collect, manage, and cite your research sources.
2. Another one is to use Mendeley.
In addition to being multi-platform, it also has
a web site where you can sync your PDFs for online access. It also extracts the metadata from your PDFs (like title,
authors, journal titles) and fills in many fields for you. You also have PDF renaming options.
3. Finally, a lot of people use papers: http://mekentosj.com/papers/
I haven't tried any of these though, so I wouldn't know which one to recommend. I personally just organize the PDF's by fields of research and name each one as the title of the paper. It helps having hard copies as well :D
Click to read the comments on this entry...3. Finally, a lot of people use papers: http://mekentosj.com/papers/
I haven't tried any of these though, so I wouldn't know which one to recommend. I personally just organize the PDF's by fields of research and name each one as the title of the paper. It helps having hard copies as well :D
Solving cryptograms in 10 simple steps
Oct 28, 2009X NDPZ FSY'Z RBYZ ZS IA ZLAKA RLAY XZ LBMMAYP.
Can you decode it? If not, let's go through some general strategies that will help :D
1. Words with one letter are either "A" or "I".
2. A very frequent 3-letter word is "THE".
3. There is an apostrophe then the letter after it is T or S. (DON'T, CAN'T, CAT'S, etc).
4. If it ends in a question mark, the first word is usually:
5. Frequent pairs of constants are TH, WH, SH or CH.
6. Frequent word endings are:
7. Short words where "ING" or "ED" are at the end sometimes have a double up on the last consonant, like "HOPPED" and "HUGGING".
8. In 2 letter words one is a vowel (or Y). Example, IF, BY, HE, AT.
9. If it's a word like [] ' [] .... then the second one is usually a D (or could be an M). Example, I'M or I'D.
10. Guess and test!!
Click to read the comments on this entry...2. A very frequent 3-letter word is "THE".
3. There is an apostrophe then the letter after it is T or S. (DON'T, CAN'T, CAT'S, etc).
4. If it ends in a question mark, the first word is usually:
WHO, WHAT, WHEN, WHERE, HOW or WHY.
5. Frequent pairs of constants are TH, WH, SH or CH.
6. Frequent word endings are:
"TION", "ENT", "ANT", "ING" "ERS", "ENS" and "ED".
7. Short words where "ING" or "ED" are at the end sometimes have a double up on the last consonant, like "HOPPED" and "HUGGING".
8. In 2 letter words one is a vowel (or Y). Example, IF, BY, HE, AT.
9. If it's a word like [] ' [] .... then the second one is usually a D (or could be an M). Example, I'M or I'D.
10. Guess and test!!
Divisibility Tricks
Oct 27, 2009
Is the number N divisible by.... 2? 3? 5?
Everyone knows the first trick:
N is divisible by 2 if its last digit is 0, 2, 4, 6, or 8 (that is, last digit is even).
Most people know the next trick:
N is divisible by 3 if the sum of the digits is also divisible by 3.
You can repeat this rule too.
For example: Is the number 93,225 is divisible by 3? Well...
Everyone knows the first trick:
N is divisible by 2 if its last digit is 0, 2, 4, 6, or 8 (that is, last digit is even).
Most people know the next trick:
N is divisible by 3 if the sum of the digits is also divisible by 3.
You can repeat this rule too.
For example: Is the number 93,225 is divisible by 3? Well...
9+3+2+2+5 = 21
And, 21 is divisible by 3, hence 93,225 is divisible by 3.
N is divisible by 4 if the last two digist form a number divisible by 4.
Let's do an example: Is the number 23894723985729316 divisible by 4? Well the last two digits is 16 and 16 is divisible by 4, so YES!
N is divisible by 5 if it ends in 0 or 5.
For 6, we just combine the rules for 2 and 3:
N is divisible by 6 if it is divisible by both 2 and 3.
For the rest, we will stick with prime divisors p. Consider multiples M of p until:
Now, to find out if a number is divisible by p, take the last digit of the number, multiply it by n, and add it to the rest of the number (OR: multiply it by (p - n) and subtract it from the rest of the number).
If you get an answer divisible by p (note that this includes 0), then the original number is divisible by p. Repeat the rule if you don't know the new number's divisibility.
Now try to see if you can come up with the rule for 7! One thing you might find interesting is the following post that discusses using a 'divisibility graph' for 7.
Click to read the comments on this entry...Let's do an example: Is the number 23894723985729316 divisible by 4? Well the last two digits is 16 and 16 is divisible by 4, so YES!
N is divisible by 5 if it ends in 0 or 5.
For 6, we just combine the rules for 2 and 3:
N is divisible by 6 if it is divisible by both 2 and 3.
For the rest, we will stick with prime divisors p. Consider multiples M of p until:
M*p+1=0 (mod 10)
We want the smallest such M.
Take
n = (Mp+1)/10
Consider n and p-n, and usually we just pick the lowest.Now, to find out if a number is divisible by p, take the last digit of the number, multiply it by n, and add it to the rest of the number (OR: multiply it by (p - n) and subtract it from the rest of the number).
If you get an answer divisible by p (note that this includes 0), then the original number is divisible by p. Repeat the rule if you don't know the new number's divisibility.
Now try to see if you can come up with the rule for 7! One thing you might find interesting is the following post that discusses using a 'divisibility graph' for 7.
How to solve the sudokube
Oct 26, 2009
The Sudoku Cube is a ripoff of the Rubik's Cube, where each face has the numbers 1-9 instead of colours. The goal:
It was created in 2006 by some guy named Jay Horowitz in Ohio. You can buy it at Barnes and Noble and some other places. In what follows we briefly describe how to solve it...
put the numbers 1-9 on each side with no repetition
It was created in 2006 by some guy named Jay Horowitz in Ohio. You can buy it at Barnes and Noble and some other places. In what follows we briefly describe how to solve it...
Of course there are lots of variations, including cubes with 4x4x4, and naming variations are Sodokube, Roxdoku. If you want to solve it you need to realize there are a few different
variations on the cube, so depending which one you have, the solution
will be slightly different.
Step 1: Familiarize yourself with solving a Rubik's Cube. If you don't know how to solve the Rubik's cube, then you will have a LOT of trouble with the Sudokube (trust me!).
Step 2: Note the centres of each face of your cube. Some cubes have 5's in all the centres, others have varying numbers. As in the Rubik's cube, these centres will be fixed points and stay in place when you do the "moves".
Step 3: If you have the Rubik's cube algorithm memorized, you should now have no trouble at all solving the cube!!
The end!
Click to read the comments on this entry...Step 1: Familiarize yourself with solving a Rubik's Cube. If you don't know how to solve the Rubik's cube, then you will have a LOT of trouble with the Sudokube (trust me!).
Step 2: Note the centres of each face of your cube. Some cubes have 5's in all the centres, others have varying numbers. As in the Rubik's cube, these centres will be fixed points and stay in place when you do the "moves".
Step 3: If you have the Rubik's cube algorithm memorized, you should now have no trouble at all solving the cube!!
The end!
List of famous mathematicians
Oct 23, 2009
Doing a school project on mathematicians? Don't pick Einstein like your mom would... pick one from the following list instead. Then google their name + wikipedia, and tada, instant math project done!
Hippias
Recorde, Robert
Ferrari, Ludovico
Viete, Francois
Mahavira
Bhaskara
Fibonacci
Stifel, Michael
Tartaglia, Niccolo
Cardano, Girolamo
Ahmes
Pythagoras
Hippocrates
Plato
Ceulen, Ludolph van
Stevin, Simon
Theaetetus
Archytas
Xenocrates
Theodorus
Napier, John
Cataldi, Pietro Antonio
Briggs, Henry
Kepler, Johannes
Oughtred, William
Bachet, Claude-Gaspar, de Meziriac
Mersenne, Marin
Ptolemy
Nicomachus of Gerasa
Theon of Smyrna
Diophantus I
Pappus
Iamblichus
Girard, Albert
Desargues, Girard
Descartes, Rene
Fermat, Pierre de
Proclus
Tsu Ch'ung-Chi
Brahmagupta
Al-Khwarizmi
Heron of Alexandria
Thabit ibn Qurra
Machin, John
Bernoulli, Nikolaus
Goldbach, Christian
Stirling, James
Euler, Leonhard
Buffon, Count Georges
Lambert, Johann
Eratosthenes
Diocles
Hipparchus
Lagrange, Joseph Louis
Brouncker, Lord William
Pascal, Blaise
Huygens, Christian
Newton, Isaac
Leibniz, Gottfried Wilhelm
Bernoulli, Johann
Wilson, John
Wessel, Caspar
Riemann, Bemard
Venn, John
Lucas, Edouard
Cantor, George
Lindemann, Ferdinand
Hilbert, David
Lehmer, D. N.
Hardy, G. H.
Ramanujan, Srinivasa
Erdos, Paul
Laplace, Pierre Simon de
Legendre, Adrien Marie
Nieuwland, Pieter
Ruffini, Paolo
Argand, Jean Robert
Gauss, Karl Friedrich
Brianchon, Charles
Binet, Jacques-Philippe-Marie
Möbius, August Ferdinand
Babbage, Charles
Laine, Gabriel
Steiner, Jakob
de Morgan, Augustus
Liouville, Joseph
Shanks, William
Catalan, Eugene Charles
Chu Shih-chieh
Pacioli, Fra Luca
Leonardo da Vinci
Aristotle
Menaechmus
Euclid
Archimedes
Nicomedes
Dürer, Albrecht
Hermite, Charles
Click to read the comments on this entry...Recorde, Robert
Ferrari, Ludovico
Viete, Francois
Mahavira
Bhaskara
Fibonacci
Stifel, Michael
Tartaglia, Niccolo
Cardano, Girolamo
Ahmes
Pythagoras
Hippocrates
Plato
Ceulen, Ludolph van
Stevin, Simon
Theaetetus
Archytas
Xenocrates
Theodorus
Napier, John
Cataldi, Pietro Antonio
Briggs, Henry
Kepler, Johannes
Oughtred, William
Bachet, Claude-Gaspar, de Meziriac
Mersenne, Marin
Ptolemy
Nicomachus of Gerasa
Theon of Smyrna
Diophantus I
Pappus
Iamblichus
Girard, Albert
Desargues, Girard
Descartes, Rene
Fermat, Pierre de
Proclus
Tsu Ch'ung-Chi
Brahmagupta
Al-Khwarizmi
Heron of Alexandria
Thabit ibn Qurra
Machin, John
Bernoulli, Nikolaus
Goldbach, Christian
Stirling, James
Euler, Leonhard
Buffon, Count Georges
Lambert, Johann
Eratosthenes
Diocles
Hipparchus
Lagrange, Joseph Louis
Brouncker, Lord William
Pascal, Blaise
Huygens, Christian
Newton, Isaac
Leibniz, Gottfried Wilhelm
Bernoulli, Johann
Wilson, John
Wessel, Caspar
Riemann, Bemard
Venn, John
Lucas, Edouard
Cantor, George
Lindemann, Ferdinand
Hilbert, David
Lehmer, D. N.
Hardy, G. H.
Ramanujan, Srinivasa
Erdos, Paul
Laplace, Pierre Simon de
Legendre, Adrien Marie
Nieuwland, Pieter
Ruffini, Paolo
Argand, Jean Robert
Gauss, Karl Friedrich
Brianchon, Charles
Binet, Jacques-Philippe-Marie
Möbius, August Ferdinand
Babbage, Charles
Laine, Gabriel
Steiner, Jakob
de Morgan, Augustus
Liouville, Joseph
Shanks, William
Catalan, Eugene Charles
Chu Shih-chieh
Pacioli, Fra Luca
Leonardo da Vinci
Aristotle
Menaechmus
Euclid
Archimedes
Nicomedes
Dürer, Albrecht
Hermite, Charles
Google results based on number of o's
Oct 13, 2009
Over on the xkcd forums there was a post about the number of google results for the word google with a varying number of "o's". Such as:
gggle
goggle
google
gooogle
...
goooooo *125 o's in total* oooogle
gggle
goggle
gooogle
...
goooooo *125 o's in total* oooogle
The data yoteango collected is as follows:
gooooooooooooooooooooooooooooooooooooooooooooooooooooooooooogle.com
is of course registered, since it's the most o's you can use. A while back, someone put a fake website up there and listed it on stumble upon, so it has a lot of links pointed towards it, giving it many more search results.
The 13 o's was also a registered domain name for some google project, hence the bump at that number. The other bumps, I'm not too sure about.
Click to read the comments on this entry...Two o's is google, so you would expect that to have the most search results. The bump at 59 is also reasonable and can be explained by the registration of the associated .com. In particular, a .com domain name can have up to 63 characters in it, and due to Google's popularity
O's Search Results----------------------------
0 105,000
1 1,500,000
2 2,020,000,000
3 342,000
4 666,000
5 50,400
6 86,500
7 33,600
8 11,300
9 8,940
10 65,400
11 11,100
12 27,300
13 107,000
14 36,000
15 11,900
16 15,000
17 45,700
18 12,100
19 948
20 2,370
21 860
22 728
23 3,030
24 8,940
25 5,990
26 2,350
27 972
28 432
29 307
30 344
31 1,970
32 1,890
33 1,530
34 1,310
35 1,470
36 11,200
37 2,140
38 187
39 1,810
40 1,070
41 1,030
52 1,010
53 1,300
44 954
45 1,210
46 963
47 786
48 666
49 1,190
50 597
51 95
52 482
53 6,140
54 677
55 453
56 785
57 711
58 466
59 3,530
60 373
61 111
62 109
63 118
64 176
65 192
66 90
67 52
68 89
69 72
70 92
71 51
72 444
73 43
74 40
75 50
76 35
77 41
78 46
79 103
80 41
81 34
82 37
83 56
84 53
85 55
86 87
87 52
88 25
89 31
90 26
91 29
92 95
93 42
94 144
95 30
96 47
97 26
98 49
99 7
100 38
101 5
102 7
103 7
104 3
105 5
106 6
107 29
108 55
109 7
110 9
111 5
112 4
113 25
114 4
115 25
116 4
117 10
118 75
119 9
120 5
121 3
122 10
123 65
124 123
gooooooooooooooooooooooooooooooooooooooooooooooooooooooooooogle.com
is of course registered, since it's the most o's you can use. A while back, someone put a fake website up there and listed it on stumble upon, so it has a lot of links pointed towards it, giving it many more search results.
The 13 o's was also a registered domain name for some google project, hence the bump at that number. The other bumps, I'm not too sure about.
The math behind low riding pants (how cool is it?)
Oct 10, 2009
In this post we learn all about the math behind low riding pants and how cool it really is.
I decided to do a study just to see what the relationship was. I interviewed 13,532 people and showed them each 6 pictures of people wearing low riding pants. I then asked them to select a "COOLNESS FACTOR" for each picture.
I decided to do a study just to see what the relationship was. I interviewed 13,532 people and showed them each 6 pictures of people wearing low riding pants. I then asked them to select a "COOLNESS FACTOR" for each picture.
The 'coolness factor' works as follows:

The results were amazing! I conclude that:
Click to read the comments on this entry...0 = super dorkyAfter gathering all of my results and putting it into this complex mathematical program and calculating means and standard deviations and something to do with chi, I found the following results:
1 = dorky
2 = average
3 = above average
4 = cool
5 = super cool!

The results were amazing! I conclude that:
- the higher the pants, the more dorky you are
- pants at waistline level means the person is average
- pants that are super low make you super cool
How to count with your fingers (efficiently?)
Oct 9, 2009
In this post I will teach you how to count using only your fingers! That's right boys and girls!!
Disclaimer: I first have to apologize to those people who have lost their hands (or some fingers), but this will still apply to you (however, you won't be able to count as high). Now if you have no hands but are good with putting your toes up and down then this will also work. If you have no hands and no feet... then ask a friend to help you. If your friend has no fingers and no toes, then ask one who DOES to help you. If you have no friends that have fingers or toes, then... monkey sticks >_>
Method 1 (Traditional Method):
Yup, you guessed it!
one finger = 1
two fingers = 2
three fingers = 3
...
nine fingers = 9
ten fingers = 10
Amazing! But did you know you can count to 10 using only one hand? The chinese are WAYYYY ahead of us (north americans) when it comes to counting. This takes us to method 2.
Disclaimer: I first have to apologize to those people who have lost their hands (or some fingers), but this will still apply to you (however, you won't be able to count as high). Now if you have no hands but are good with putting your toes up and down then this will also work. If you have no hands and no feet... then ask a friend to help you. If your friend has no fingers and no toes, then ask one who DOES to help you. If you have no friends that have fingers or toes, then... monkey sticks >_>
Method 1 (Traditional Method):
Yup, you guessed it!
one finger = 1
two fingers = 2
three fingers = 3
...
nine fingers = 9
ten fingers = 10
Amazing! But did you know you can count to 10 using only one hand? The chinese are WAYYYY ahead of us (north americans) when it comes to counting. This takes us to method 2.
Method 2 (Chinese Number Gestures):
If you are chinese you may already know this method. I have many chinese friend and they teach me counting. The idea is you can count from 0 to 10 using only ONE hand. The best way to describe it is in picture form. The numbers from 0 to 5 are easy and exactly what you would expect:

That was easy, wasn't it? What about the other numbers? Well not so easy, but once you learn it you never forget it (well I actually forget it but i'm not as smart as you). Let's begin:

The last one is an alternative for 10 using two hands. Neato! Now go show all your friends.. go go go! Oh wait, there's more, so hold on, don't go yet... read the rest of this post... Onto method 3.
Method 3: Finger Binary
This method lets you count from 0 to 1023 with two hands. For those of us who are mathematically inclined, you can probably guess how this method works. If you don't know monkey sticks about math then I am here to HELP! Binary means 0 or 1.
In binary, the far right digit represents 2^0, the one to the left of it 2^1, then 2^2, and so on. For example,
100101 = 1 x 2^5 + 0 x 2^4 + 0 x 2^3 + 1 x 2^2 + 0 x 2^1 + 1 x 2^0 = 37
0 in binary is 0
1 in binary is 1
2 in binary is 10 (that is, we have one 2^1 and zero 2^0's totaling 2)
3 in binary is 11 (that is, we have one 2^1 and one 2^0's making 3)
4 in binary is 100
5 in binary is 101
6 in binary is 110
7 in binary is 111
and so on...
With our fingers representing 0's and 1's, we can represent a binary digit with 10 columns:
* * * * * * * * * *
The largest being
1111111111 (which in normal numbers is 1023).
Now the standard way to do this is by the following table:
The values of each raised finger are then added together to get the total number. Okay, let's do some picture examples!

Can you guess what this is? The person is using their RIGHT hand and has the pinky, ring and index fingers up. If you know binary you can just compute the number easily, if not, use the above table to get:

Click to read the comments on this entry...If you are chinese you may already know this method. I have many chinese friend and they teach me counting. The idea is you can count from 0 to 10 using only ONE hand. The best way to describe it is in picture form. The numbers from 0 to 5 are easy and exactly what you would expect:

That was easy, wasn't it? What about the other numbers? Well not so easy, but once you learn it you never forget it (well I actually forget it but i'm not as smart as you). Let's begin:
- Six: Thumb and pinky extended.
- Seven: Umm, I don't know how to describe.. umm... it's like you hand just ate something and has it's mouth closed.
- Eight: It's L-shape with thumb and index.
- Nine: It looks like a hook to me.
- Ten: This is like "please please please" (fingers crossed).

The last one is an alternative for 10 using two hands. Neato! Now go show all your friends.. go go go! Oh wait, there's more, so hold on, don't go yet... read the rest of this post... Onto method 3.
Method 3: Finger Binary
This method lets you count from 0 to 1023 with two hands. For those of us who are mathematically inclined, you can probably guess how this method works. If you don't know monkey sticks about math then I am here to HELP! Binary means 0 or 1.
In binary, the far right digit represents 2^0, the one to the left of it 2^1, then 2^2, and so on. For example,
100101 = 1 x 2^5 + 0 x 2^4 + 0 x 2^3 + 1 x 2^2 + 0 x 2^1 + 1 x 2^0 = 37
0 in binary is 0
1 in binary is 1
2 in binary is 10 (that is, we have one 2^1 and zero 2^0's totaling 2)
3 in binary is 11 (that is, we have one 2^1 and one 2^0's making 3)
4 in binary is 100
5 in binary is 101
6 in binary is 110
7 in binary is 111
and so on...
With our fingers representing 0's and 1's, we can represent a binary digit with 10 columns:
* * * * * * * * * *
The largest being
1111111111 (which in normal numbers is 1023).
Now the standard way to do this is by the following table:
| . | | | Left Hand | | | Right Hand | ||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|
| ... | | | Thumb | Index | Middle | Ring | Pinky | | | Pinky | Ring | Middle | Index | Thumb |
| Power | | | 2^9 | 2^8 | 2^7 | 2^6 | 2^5 | | | 2^4 | 2^3 | 2^2 | 2^1 | 2^0 |
| Value | | | 512 | 256 | 128 | 64 | 32 | | | 16 | 8 | 4 | 2 | 1 |
The values of each raised finger are then added together to get the total number. Okay, let's do some picture examples!

Can you guess what this is? The person is using their RIGHT hand and has the pinky, ring and index fingers up. If you know binary you can just compute the number easily, if not, use the above table to get:
16 + 8 + 2 = 26
WOW! AMAZING! Now go tell all your friends! And
the next time you have to count to 874 try to use the binary method
(it'll really get you thinking and you'll become an expert binarician). Let's do one more example to make sure we understand:
Notice how he has his RIGHT hand up and only his middle finger. His left hand doesn't have any fingers up. Thus, using the above table we get FOUR! Easy wasn't it? So remember kids, the next time you want four of something, use "Binary Finger Notation" as shown by the gentleman above!
The mathematics of sudoku
Oct 7, 2009
Tom Davis has a great article on the mathematics of sudoku. He first describes a brief history of the puzzle and how to play. Then he discusses why it is mathematically interesting (it is, trust me!). He goes through some obvious strategies that a lot of people try when doing sudoku and some other clever strategies. Definitely check it out if you have time!
Click to read the comments on this entry...
Math fails in history
Oct 3, 2009
Dick Lipton wrote a great post over at Gödel's Lost Letter and P=NP. In his Sept 27th post he talked about surprises in mathematics. In one of his sections he gives three examples of where mathematicians "accepted" a false proof. Sometimes this happens and it might be dozens of years until someone realizes a mistake has been made.
One interesting example of this is the Four Colour Theorem (that's right ya bunch of monkeys, I spelled colour with a U!!!)...
One interesting example of this is the Four Colour Theorem (that's right ya bunch of monkeys, I spelled colour with a U!!!)...
Lipton says...
Click to read the comments on this entry..."The Four-Color Theorem (4CT) dates back to 1852, when it was first proposed as a conjecture. Francis Guthrie was trying to color the map of counties in England and observed that four colors were enough. Consequently, he proposed the 4CT. In 1879, Alfred Kempe provided a "proof" for the 4CT. A year later, Peter Tait proposed another proof for 4CT. Interestingly both proofs stood for 11 years before they were proved wrong. Percy Heawood disproved Kempe's proof in 1890, and Julius Petersen showed that Tait's proof was wrong a year later.Go check out the rest of his post NOW.
However, Kempe's and Tait's proofs, or attempts at a proof, were not fully futile. For instance, Heawood noticed that Kempe's proof can be adapted into a correct proof of a "Five-Color Theorem". There were several attempts at proving the 4CT before it was eventually proved in 1976. See this article by Robin Thomas for a historical perspective of the problem."
How many guys are there for me?
Sep 29, 2009Hey guys,
Based on my previous post of "How many girls are there for me," I thought it would be interesting to see how many guys there are for me. So hypothetically let's pretend I am gay and a top (if you don't know what that means you should get some more gay friends lol). Just a recap that I found out in the last post that:
Now, some of my stats I used from counting the number of girls for me still hold. The male population on the planet is about:
Based on my previous post of "How many girls are there for me," I thought it would be interesting to see how many guys there are for me. So hypothetically let's pretend I am gay and a top (if you don't know what that means you should get some more gay friends lol). Just a recap that I found out in the last post that:
11, 414 girls
are suitable matches for me.Now, some of my stats I used from counting the number of girls for me still hold. The male population on the planet is about:
Male population on earth in 2009:
3,500,000,000
3,500,000,000
But I am looking for a boy who is in North America. Restricting this number to Canada and the United States gives:
He must also be around my age. Say, from 25-29 years old. That leaves:
But about half of these boys are already married or in a common-law relationship. Thus, that leaves:
And further, about 89% of men are not interested in men (i.e. straight or other). That leaves:
Being a mathematician, I want a boy who is smart. But a lot of gay guys are actually smart (unlike the other post which quoted: 85% of girls in North America are complete dumb asses). Only about 50% of gay guys are complete dumbasses, thus that leaves:
But I don't want no ugly man! Thank god that a lot of gay guys take care of their appearance and work out at the gym. About 40% of gay guys don't, so that leaves:
But the boy must also like me. To see what kind of statistic is reasonable, I went to the local gay bar and chatted with 10 boys. Then at the end I asked them if they liked me based on my appearance and our conversation. An outstanding 7 boys said yes and wanted to go back to my place. But upon further asking, 5 of them didn't want an actual relationship with me. So, only about 20% of the boys actually like me and would date me. That leaves:
Finally, as I said at the top, hypothetically speaking I would be a "top" gay. So, assuming that 50% are also tops, that would leave me with:
Holy cow!! WTF!? So only 11, 414 girls would make a perfect match with me, but if I were gay, an outstanding 18, 834 boys would make a perfect match with me! Damn, I better reconsider my options LOL.
Click to read the comments on this entry...171, 230, 000 boys
He must also be around my age. Say, from 25-29 years old. That leaves:
11, 415, 000 boys
But about half of these boys are already married or in a common-law relationship. Thus, that leaves:
5, 707, 500 boys
And further, about 89% of men are not interested in men (i.e. straight or other). That leaves:
627, 825 boys
Being a mathematician, I want a boy who is smart. But a lot of gay guys are actually smart (unlike the other post which quoted: 85% of girls in North America are complete dumb asses). Only about 50% of gay guys are complete dumbasses, thus that leaves:
313, 912 boys
But I don't want no ugly man! Thank god that a lot of gay guys take care of their appearance and work out at the gym. About 40% of gay guys don't, so that leaves:
188, 347 boys
But the boy must also like me. To see what kind of statistic is reasonable, I went to the local gay bar and chatted with 10 boys. Then at the end I asked them if they liked me based on my appearance and our conversation. An outstanding 7 boys said yes and wanted to go back to my place. But upon further asking, 5 of them didn't want an actual relationship with me. So, only about 20% of the boys actually like me and would date me. That leaves:
37, 669 boys
Finally, as I said at the top, hypothetically speaking I would be a "top" gay. So, assuming that 50% are also tops, that would leave me with:
18, 834 boys
Holy cow!! WTF!? So only 11, 414 girls would make a perfect match with me, but if I were gay, an outstanding 18, 834 boys would make a perfect match with me! Damn, I better reconsider my options LOL.
How many girls are there for me?
Sep 29, 2009
Hey guys,
I don't have a girlfriend but I don't think there is anything wrong with me, hahaha... but being the mathematician that I am, let's look to see how many girls exactly would be a match for me :D
I went onto Wolfram|Alpha and looked up the female population on the planet and got:
But I am looking for a girl who is in North America. Restricting this number to Canada and the United States gives:
I don't have a girlfriend but I don't think there is anything wrong with me, hahaha... but being the mathematician that I am, let's look to see how many girls exactly would be a match for me :D
I went onto Wolfram|Alpha and looked up the female population on the planet and got:
Female population on earth in 2009:
3,326,000,000
3,326,000,000
But I am looking for a girl who is in North America. Restricting this number to Canada and the United States gives:
171, 230, 000 girls
BUT she must also be around my age and preferably younger (or at most 1 year older). Say, from 25-29 years old. That leaves:
But about half of these girls are already married or in a common-law relationship. Thus, that leaves:
And further, about 11% of women are not interested in men (i.e. lesbian or other). That leaves:
Being a mathematician, I want a girl who is smart. About 85% of the girls in North America are complete dumb asses, hence the number of girls left is:
But I don't want no ugly kids, and given my appearance, the girl better be pretty! Only about 15% of the girls fall in this category, thus we get:
But the girl must also like me. To see what kind of statistic is reasonable, I went to the local bar and chatted with 10 girls. Then at the end I asked them if they liked me based on my appearance and our conversation. Only 1 girl said yes, the other 9 said no (unfortunately the 1 girl who said yes went home with my friend :-S). Thus, only 10% of the girls will like me. That leaves:
Further, I have two sisters and a few other relatives who would fall into the above categories, and who I prefer not to marry. So taking away these 10-15 people gives:
WOW! So in this world there is 11,414 girls that would like me and would marry me. That's actually a lot (in my mind). Now if only there is a way to meet them, hmmm.
I thought it would also be very interesting to see how many GUYS there are for me and compare... you can read that calculation at my next entry:
Click to read the comments on this entry...11, 415, 000 girls
But about half of these girls are already married or in a common-law relationship. Thus, that leaves:
5, 707, 500 girls
And further, about 11% of women are not interested in men (i.e. lesbian or other). That leaves:
5, 079, 675 girls
Being a mathematician, I want a girl who is smart. About 85% of the girls in North America are complete dumb asses, hence the number of girls left is:
761, 951 girls
But I don't want no ugly kids, and given my appearance, the girl better be pretty! Only about 15% of the girls fall in this category, thus we get:
114, 292 girls
But the girl must also like me. To see what kind of statistic is reasonable, I went to the local bar and chatted with 10 girls. Then at the end I asked them if they liked me based on my appearance and our conversation. Only 1 girl said yes, the other 9 said no (unfortunately the 1 girl who said yes went home with my friend :-S). Thus, only 10% of the girls will like me. That leaves:
11, 429 girls
Further, I have two sisters and a few other relatives who would fall into the above categories, and who I prefer not to marry. So taking away these 10-15 people gives:
11, 414 girls
WOW! So in this world there is 11,414 girls that would like me and would marry me. That's actually a lot (in my mind). Now if only there is a way to meet them, hmmm.
I thought it would also be very interesting to see how many GUYS there are for me and compare... you can read that calculation at my next entry:
All winners who are in Canada have to answer math problems before getting any prize. Things like...
This question was for claiming an Xbox 360 in a contest, so the answer of course is 360!
But WHY?? Americans don't have to do it, a monkey on a stick can win prize in the U.S and not even know how to do 1+2. But in Canada, to even win the McDonald's hockey trading cards you need to do these silly computations.
90 x 2, then divide by six, then x 12
This question was for claiming an Xbox 360 in a contest, so the answer of course is 360!
But WHY?? Americans don't have to do it, a monkey on a stick can win prize in the U.S and not even know how to do 1+2. But in Canada, to even win the McDonald's hockey trading cards you need to do these silly computations.
This is actually a loop-hole in Canadian law. Basically, Canada has
anti-gambling laws that make it illegal to sell chances to win a prize,
so there has to be at least some level of skill involved. In the past,
the "skill" challenges involved things like:
In 1984 is when something big happened. A court case said that a four-part mathematical question counts as a "test of skill". Something like:
However, in the last 20 years, the four-part math problem has turned into monkey sticks... some have simplified it so much that the division they require is by 1. And forget about using 3 digit numbers, 2 digits is good enough. And four-part problem? BAH i say! BAH! Three-parts should be enough, right? So now you see things like:
Click to read the comments on this entry...- Count the number of jelly beans in a jar
- Calculate the time it takes for a barrel to float down a river
- Shoot a turkey at 50 yard range
- Quickly peel a potato
In 1984 is when something big happened. A court case said that a four-part mathematical question counts as a "test of skill". Something like:
228 x 21, then add 10824, then divide by 12, then subtract 1121.
Due
to this court case, almost all product promotion sweepstakes started
using the four-part math problem to test skill and get around the nasty
Canadian laws. However, in the last 20 years, the four-part math problem has turned into monkey sticks... some have simplified it so much that the division they require is by 1. And forget about using 3 digit numbers, 2 digits is good enough. And four-part problem? BAH i say! BAH! Three-parts should be enough, right? So now you see things like:
2 x 2, then add 10, then divide by 1.
But is the Canadian government going to crack down on companies posing super easy skill-testing questions? Probably not...
Girls are dumb at math
Sep 26, 2009
Oh jeez, with a title like that I'm sure to get my ass kicked :-\
"In tests in Canada, women who were told that men and women do math equally well did much better than those who were told there is a genetic difference in math ability.
And women who heard there were differences caused by environment -- such as math teachers giving more attention to boys -- outperformed those who were simply reminded they were females.
The women who did better in the tests got nearly twice as many right answers as those in the other groups, explained Steven J. Heine, a psychology professor at the University of British Columbia in Vancouver."
The math study was done a couple years ago from 2003-2006. Heine and
Dar-Nimrod wanted to see how people are affected by stereotypes about
themselves. They took 220 women and divided them into 4 groups and gave
them math and reading tests. The results were published in the journal
Science.
The way they did the experiment was to give the women a math test, then have them read an essay, then give them another math test. In two groups the women averaged between five and 10 correct answers out of 25 math questions. In the other two they averaged between 15 and 20 correct.
The women in the lower socring groups read essays about the 'genetic difference between men and women in math ability' or read an essay about the 'images of women in art' (which reminds them they are female but didn't discuss math). These two groups had a decreased performance between the two math tests. Thus... reminding people of the stereotype affects them.
Click to read the comments on this entry...The way they did the experiment was to give the women a math test, then have them read an essay, then give them another math test. In two groups the women averaged between five and 10 correct answers out of 25 math questions. In the other two they averaged between 15 and 20 correct.
The women in the lower socring groups read essays about the 'genetic difference between men and women in math ability' or read an essay about the 'images of women in art' (which reminds them they are female but didn't discuss math). These two groups had a decreased performance between the two math tests. Thus... reminding people of the stereotype affects them.
Identical lottery draw
Sep 18, 2009
In Bulgaria's national lottery, the same six winning numbers were drawn twice in a row. Minister Svilen Neikov ordered an investigation after the
numbers 4, 15, 23, 24, 35 and 42 were selected, in a different order,
by a machine live on television on September 6th and 10th. Some thought the results were manipulated, however after an investigation there was no wrongdoing found.
A total of 18 people got all six numbers when they were drawn the second time and each got $7,700. Nobody guessed right the first draw.
Mathematicians say that the chance of drawing the same six numbers in two consecutive rounds is about 1 in 4.2 million.
The following website has some calculations dealing with lottery math and whether it's better to play 50 dollars in one lottery, or play one dollar in fifty lotteries.
Click to read the comments on this entry...A total of 18 people got all six numbers when they were drawn the second time and each got $7,700. Nobody guessed right the first draw.
Mathematicians say that the chance of drawing the same six numbers in two consecutive rounds is about 1 in 4.2 million.
The following website has some calculations dealing with lottery math and whether it's better to play 50 dollars in one lottery, or play one dollar in fifty lotteries.
When to get married
Sep 14, 2009
Do you know when to get married?
When you meet someone, how do you know you won't find someone better? Love aside, let's look at this from a mathematical viewpoint.
Suppose you are a guy, and are interested in say... girls. Let's say over the course of your life you meet 100 girls and each girl is ranked by how good of a mate she would be with you. The obvious solution is to just meet each of the 100 girls and pick the one that is ranked the highest. BUT... by the time you met all 100, you will be an old geezer and on your death bed. So what is one to do?
When you meet someone, how do you know you won't find someone better? Love aside, let's look at this from a mathematical viewpoint.
Suppose you are a guy, and are interested in say... girls. Let's say over the course of your life you meet 100 girls and each girl is ranked by how good of a mate she would be with you. The obvious solution is to just meet each of the 100 girls and pick the one that is ranked the highest. BUT... by the time you met all 100, you will be an old geezer and on your death bed. So what is one to do?
Mathematicians have already worked out all the details. They say that
the best choice is to wait until meeting 37 of the 100 girls (more
specifically: 36.7879441). So just more than a third of the way into
your dating life you should get married. [For your information, the
number 0.367879441 is actually 1/e.]
So assuming that the dating range for men is like... 20 to 50, a man had about 30 years to find a wife. Since 30/e ~= 11, that means a man should get married around the age of 31. By that age, he would have met a girl who ranks high enough in his books :-D.
Considering most of my friends got married before they were 21, I'd say they definitely weren't worried about the mathematics behind it.
Source: John Gilbert and Frederick Mosteller, Recognizing the maximum of a sequence. J. Amer. Statist. Assoc. 61 1966 35--73.
Link: http://www.jstor.org/stable/2283044?seq=1
Click to read the comments on this entry...So assuming that the dating range for men is like... 20 to 50, a man had about 30 years to find a wife. Since 30/e ~= 11, that means a man should get married around the age of 31. By that age, he would have met a girl who ranks high enough in his books :-D.
Considering most of my friends got married before they were 21, I'd say they definitely weren't worried about the mathematics behind it.
Source: John Gilbert and Frederick Mosteller, Recognizing the maximum of a sequence. J. Amer. Statist. Assoc. 61 1966 35--73.
Link: http://www.jstor.org/stable/
Happy 09/09/09 day!
Sep 9, 2009
For those of you who don't know, it is indeed 09/09/09 day. Woohoo? Ya I know, it's just another day, but for some reason the general public LOVES symmetry, so let's give them what they like to hear ;-)
First, since it's not a leap year, September 9, 2009, is the 252nd day of the year. But,
Second,
First, since it's not a leap year, September 9, 2009, is the 252nd day of the year. But,
2+5+2=9
Second,
9 x 9 x 9 = 27
and2+7=9
WOW! Are you amazed yet (directed towards the non-mathematicians). But wait, there's more!
The day falls on a Wednesday, and both Wednesday and September have:
A nice fact is that 9 in mandarin (chinese) sounds like forever. So 999 emphasizes this and is considered very lucky. So don't be surprised if there are a lot of weddings today.
But wait, there's more! 09/09/09 is also the last of single digit dates for a long time. Actually, 92 years until it happens again. And guess what?
Boy, I just can't wait until 10/10/10 day!
Click to read the comments on this entry...9 letters!
Double
WOW! And of course if you put 999 upside down you get the number people
associate with evil. Because of this, some people think the world is
going to end today (it's not so take off your tinfoil hats please).A nice fact is that 9 in mandarin (chinese) sounds like forever. So 999 emphasizes this and is considered very lucky. So don't be surprised if there are a lot of weddings today.
But wait, there's more! 09/09/09 is also the last of single digit dates for a long time. Actually, 92 years until it happens again. And guess what?
9+2=11
Neat eh? And also,1+1=2
and there is a 2 in 92, and also 27. And 9^2 is 81. And there is a 1 in both 11 and 81!! WOW! Okay okay, I'll stop. ^_^Boy, I just can't wait until 10/10/10 day!
How to Turn a Sphere Inside Out Video
Aug 20, 2009
This cool youtube video shows how to turn a sphere inside out without making a hole, tearing it, or creasing it!

Click to read the comments on this entry...
How to Compute Cubed Roots Fast
Aug 9, 2009
Take a look at this video of Scott Flansburg on the Discovery Channel's "More Than Human":


In the video you see Scott Flansburg take the cubed root of 658,503 to get an answer of 87 in a matter of a second. How does he do it you ask?
This trick does require some memorization though, and also requires the
number given to be a perfect cube. You need to memorize the cubes of
the numbers 0 through 9 (or be able to figure them out on the spot).
This information is contained below:

Note that the last digits of the cubes on the right have all the numbers 1 to 9, but no number is repeated. Here is how to find the two-digit cube root of a perfect cube.
Take a number, such as 658,503 which is grouped into two parts.
1. Looking at the number we see it ends in a 3, and according to the table only 7^3 ends in a 3, thus the last digit of our number is 7.
2. Next, ignore the last 3 digits of the cube, so consider 658. Compare these digits with the table above. Note that 658 fits between 512 and 729. You always choose the smaller one, in this case 512 which happens to correspond to 8^3.
Thus, the last digit is 7 and the first digit is 8, giving an answer of 87.
Normally this trick is used for six digit perfect cubes. To help understand how this works, ask yourself - What is the last digit of (10x+y)^3? Clearly it is y^3 mod 10 (how does this relate to #1?).
Another Example:
In 474,552 we have that 343 is the immediate smallest number from 474 so the first digit is 7.
The last digit in 474,552 is 2 and only 8^3 ends in a 2, so the last digit is 8. Hence, 78^3=474,552.
Click to read the comments on this entry...
Note that the last digits of the cubes on the right have all the numbers 1 to 9, but no number is repeated. Here is how to find the two-digit cube root of a perfect cube.
Take a number, such as 658,503 which is grouped into two parts.
1. Looking at the number we see it ends in a 3, and according to the table only 7^3 ends in a 3, thus the last digit of our number is 7.
2. Next, ignore the last 3 digits of the cube, so consider 658. Compare these digits with the table above. Note that 658 fits between 512 and 729. You always choose the smaller one, in this case 512 which happens to correspond to 8^3.
Thus, the last digit is 7 and the first digit is 8, giving an answer of 87.
Normally this trick is used for six digit perfect cubes. To help understand how this works, ask yourself - What is the last digit of (10x+y)^3? Clearly it is y^3 mod 10 (how does this relate to #1?).
Another Example:
In 474,552 we have that 343 is the immediate smallest number from 474 so the first digit is 7.
The last digit in 474,552 is 2 and only 8^3 ends in a 2, so the last digit is 8. Hence, 78^3=474,552.
How to Guess the Number of M&Ms in a Jar
Aug 6, 2009
Adam Micolich has posted a video to show how using the "packing fraction" one can come up with a very accurate guess to the number of M&Ms in a jar.

Click to read the comments on this entry...
Using Fibonacci for mile <--> km conversion
Jul 29, 2009
The Fibonacci sequence:
1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ...
can be used to convert miles to kilometers (and vice versa). For example, to find how many km 5 miles is, take the next Fibonacci number which happens to be 8. Thus, 5 miles is approximately 8 km. Similarly, 8 miles is approximately 13 km, and so on.
This works because the growth rate of the Fibonacci numbers converges to the golden ratio (approx 1.618) which happens to be very close the km/mile conversion (1 mile = 1.609 km).
Click to read the comments on this entry...1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ...
can be used to convert miles to kilometers (and vice versa). For example, to find how many km 5 miles is, take the next Fibonacci number which happens to be 8. Thus, 5 miles is approximately 8 km. Similarly, 8 miles is approximately 13 km, and so on.
This works because the growth rate of the Fibonacci numbers converges to the golden ratio (approx 1.618) which happens to be very close the km/mile conversion (1 mile = 1.609 km).
Prime number trick (with solution)
Jul 22, 2009
Using prime numbers, you can amaze your friends with a prime prediction...
1. Ask your friends to pick any prime number greater than 3.
2. Square it.
3. Add 14.
4. Divide by 12.
Without knowing which prime number your friends picked, you can still tell them:
There will be a remainder of 3.
But HOW does it work?
1. Ask your friends to pick any prime number greater than 3.
2. Square it.
3. Add 14.
4. Divide by 12.
Without knowing which prime number your friends picked, you can still tell them:
There will be a remainder of 3.
But HOW does it work?
Let's do an example:
13 is a prime number, squaring it gives 169, adding 14 gives 183 which has a remainder of 3 upon division by 12.
This works for every prime number greater than 3, but how exactly does it work?
The mathematics behind this is rather simple.
1. Let p be a prime number, p > 3.
2. Squaring gives:
p^2.
3. Adding 14 gives:
p^2 + 14
4. Taking it modulo 12 gives:
(p^2 + 14) mod 12
We want to show that:
(p^2 + 14) mod 12 = 3
This is equivalent to:
p^2 - 1 is divisible by 12.
That is:
(p-1)(p+1) is divisible by 12.
For a number to be divisible by twelve, it has to be divisible both by 3 and by 4. We know that, out of p-1, p and p+1, one of them must be divisible by 3; and it can't be p, because p is prime and greater than 3. Thus, either p-1 or p+1 is divisible by 3, and so their product is also:
(p-1)(p+1) is divisible by 3.
Now, since p is a prime greater than 3, we know that it is odd. Therefore, both p-1 and p+1 are even numbers. The product of two even numbers is divisible by 4, so:
(p-1)(p+1) is divisible by 4.
Combining this with the above, we get that:
(p-1)(p+1) is divisible by 12.
And hence:
(p^2 + 14) mod 12 = 3
Click to read the comments on this entry...13 is a prime number, squaring it gives 169, adding 14 gives 183 which has a remainder of 3 upon division by 12.
This works for every prime number greater than 3, but how exactly does it work?
The mathematics behind this is rather simple.
1. Let p be a prime number, p > 3.
2. Squaring gives:
p^2.
3. Adding 14 gives:
p^2 + 14
4. Taking it modulo 12 gives:
(p^2 + 14) mod 12
We want to show that:
(p^2 + 14) mod 12 = 3
This is equivalent to:
p^2 - 1 is divisible by 12.
That is:
(p-1)(p+1) is divisible by 12.
For a number to be divisible by twelve, it has to be divisible both by 3 and by 4. We know that, out of p-1, p and p+1, one of them must be divisible by 3; and it can't be p, because p is prime and greater than 3. Thus, either p-1 or p+1 is divisible by 3, and so their product is also:
(p-1)(p+1) is divisible by 3.
Now, since p is a prime greater than 3, we know that it is odd. Therefore, both p-1 and p+1 are even numbers. The product of two even numbers is divisible by 4, so:
(p-1)(p+1) is divisible by 4.
Combining this with the above, we get that:
(p-1)(p+1) is divisible by 12.
And hence:
(p^2 + 14) mod 12 = 3



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