Saturday, September 24, 2011

Check out my Mathemagics Performance, and even learn a trick!!!

I don't know if you have seen my performances before, so today is your opportunity to see it. Basically, I spend the show doing very big multiplication and squaring problems, create magic squares, tell people the day of the week they were born on, and more. Here is a compilation of some of my performances.

If you liked it, you can see me live at the Chicago Toy and Game Fair in November, and many other possible shows. These are listed on my website,

In this video, you saw me square numbers. This is a method like the close-together method, but even easier than squaring numbers, and even the things I've taught you all along is squaring a two-digit number ending in five. A crowd of fifth graders could understand it, so hopefully, you guys are smarter than those fifth graders!!

Let's take the number 35. In order to square it, you need to remember two things:

1) The last two digits are always 25.
2) The start of the number is the first digit times the first digit plus one.

So, first, we take the first digit, or 3. Three times four (3 + 1) is equal to 12. Since the last two digits are always 25, 35^2 is 1225.

Let's look at, say 85^2. 8 x 9 = 72, so the answer is 7225. Very simple, right.

Let me make it a tad harder on you. Try 145^2. All we need to do is think of it as a two-digit number, with fourteen as the first digit and five as the last one. So, 14 x 15 isn't too bad, as you remember from the first post. If you try it yourself, you will get the number 210. Since it always ends in 25, the answer is 21025.

With very little practice, you will be able to square two-digit numbers ending in five. If you practice the close-together method I taught on August 20, you will be able to square three-digit numbers ending in five. Have fun!!

August Problem of the Week Answers:


b = 35
Explicit Formula: n - 1
Recursive Formula: an-1 + 1
g = 34
a = 908
p = 5%


z = 42.5
a = 1/2
b = 1/2
c = 1
y = 55
x = 10

Saturday, September 17, 2011

What does .9999999999999... really mean?

Haven't we all heard people say that there is a 99.999 "repeating" percent chance that something will happen. They think that this means they are almost sure it will happen. Though I don't want to make things complicated, this is actually equal to 100 percent.

To make things more simple, let's look at if 0.9999... = 1. First off, realize that 1/3 x 3 = 1, but .33333... x 3 = .99999... Same with 1/7, or other repeating fractions.

However, this won't convince people. So, I like to show them my favorite way to look at it, the algebraic proof, the one that proves this fact, or all of algebra and other mathematics incorrect. Say that .9999... is equal to S.

S = .9999...

Try multiplying that all by ten.

10S = 9.9999...
     S =    .9999...

What if we subtract both of these equations from each other.

10S = 9.9999...
   -S =    .9999...
  9S = 9
     S = 1

By dividing both sides by nine, it shows that S is both of these numbers.

This is another thing that is absolutely shocking. Before I saw this, I always thought that it was too close to call, but not exactly one. This is probably one of my favorite proofs in number theory, and all of mathematics.

Saturday, September 10, 2011

Another Probability Paradox: What's your birthday?

As I said about a month and a half ago, probability is a topic that people have difficulty understanding. I showed you the Monty Hall Paradox, where the average person is almost 17% off on the odds! 17% makes a big difference in what you should do in that scenario! Let's look at another thing that completely fools people, the Birthday Paradox.

Say you walk into a room of 23 people. What do you think the odds are that two of the people have the same birthday? Maybe like 23 in 365 because there are 365 days in a year, being about 6.3%. Would you be surprised if the true answer was over 50%?

The mistake people make is they try to determine the odds that someone has the same birthday as them. This is not correct, as there is no specification in the problem as to which two people it is. If you were to calculate those odds, it would be around 0.14%, which is nowhere close to the true odds.

In the Monty Hall Problem, I showed you how we know this. For this problem, we don't need much proof, because we can figure out the odds. Not of this question, but of the opposite question.

First off, what if we want to make a list of how many possibilities there are for 23 people. There are 365 for the first person, 365 for the second, and so on. Basically, there are a total of about 85 octodecillion (8.5 x 10^58), or 365^23. Basically, a list for n people is 365^n.

Now let's cover the rest of it. How many of these different possibilities have all different birthdays? Well, the first person's birthday could be any of 365 days. Since the second must be different, it only can choose between 364 days. The third only has 363, all the way up to the 23rd, who has 343 different possibilities. For n people, this is equal to:

365 x 364 x 363 x ... x (367 - n) x (366 - n)

You can very easily shorten this expression. On your calculator, there may be a button that is an x with an exclamation point after it. All this does is takes the number you type in and multiply it by every single whole number before it. For instance, 6! is 720 because 6 x 5 x 4 x 3 x 2 x 1 = 720. So, this expression is equal to:


And this is out of a list of 365^n possibilities. Therefore, the total possibilities there are for no one having the same birthday is:


If you subtract this decimal from one, you will know the probability that two people have the same birthday. For 23 people, we 365!/(365^n)(365-n)!]

This is about a 50.7% chance. For 30 people, we are already at 70%. 50 people is already at a 97% chance. 100 people is like a 99.99996% chance that two people will have the same birthday.

Saturday, September 3, 2011

Fibonacci Day: Adding the Odds

Today is another Fibonacci day. It is the 3rd, which is a Fibonacci number. We've looked at a bunch of addition patterns in the Fibonacci numbers, like adding them, or the even ones. What about the odds? How about we look.

1 = 1
1 + 2 = 3
1 + 2 + 5 = 8
1 + 2 + 5 + 13 = 21

See the pattern? We are getting the Fibonacci numbers! Why? Let's look at each one as the previous two.

1 + (1 + 1) + (2 + 3) + (5 + 8)

You are basically adding the Fibonacci numbers up, and then adding one! that's a Fibonacci number minus one plus one, or a plain Fibonacci number.

Bonus: We've done quite a bit with explicit formulas lately. I think now is the time to mention the explicit formula for Fibonacci numbers. For the nth Fibonacci number, you do:
[1/(5^.5)](φ^n - φ^n)

If you don't know, the Greek letter fi means the golden ratio, or 1.618... Fi with a bar on top is what you get with a small change in the golden ratio's formula, -0.618...

This is definitely complicated, but I couldn't believe that worked. I checked many numbers just to convince myself it worked!!