Math Awareness Month Part 2: Infinite Series

Today's page for Math Awareness Month is about a recent video that caused some huge debate. I saw the video a month or two ago, and was very intrigued. I showed it to some of my friends, and we were arguing about the content for quite a while. It also spread rapidly around the math department at Andover, with some teachers bringing up in their classes.

Take a look at the page and try some of the exercises. You will find the outcomes very interesting and mind-boggling. The concept of infinity is difficult for any human being to grasp, making it tons of fun to think about.

http://www.mathaware.org/mam/2014/calendar/infinity.html

Comment below what you think of the video. Do you think it is accurate? What do you think the fallacies are? How could this be a part of string theory if it is mathematically flawed?

In math class last week, we were given the following problem:

I then did the math and determined that the limit would be -1/12. I then called my teacher over, and pointed to that answer. Recalling the video, I asked him if I could rewrite that -1/12 as 1+2+3+4+5+6+7+... as my final answer. Thankfully, he got the reference. In addition to being a funny anecdote, the fact that people got the joke shows how wide of an audience this information has reached and captivated, which is amazing to see.

Figure Out The Number Of Digits In Gigantic Numbers

In science and math, you often run across numbers that are too big to be written in standard form. They are usually written in scientific notation, but they are also sometimes written as a number to a certain power. For instance, one might say that there are 2²⁰ outcomes of the flipping of twenty coins rather than saying 1.049x10⁶ ways.

By using that power, your information is likely more accurate. However, this power does not tell you much about the number. Most of us would have no idea if 2²⁰ is in the thousands, millions, billions, etc. at the first glance.

First, lets ask a question. What is the common logarithm of a number, or what can you gather from it? Well, the common logarithm is the power that ten has to be raised to to obtain that number. For instance:

log(100) = 2

log(5000) = 3.69897

log(6283185) = 6.79818

What do you notice about these numbers? It's not clear at first, but count the number of digits in each of the inputs. You will find that the common log is always just a little bit below that number. In fact, to figure out the number of digits in a number, all you have to do is take the common log and round up to the nearest integer.

How can this be used to find the number of digits in a power? Interestingly enough, there is a logarithmic identity stating that the log of a number raised to the power is equal to the power times the log of the number. For example,

log(27) = 3log(3)

ln(32) = 5ln(2)

log(2²⁰) = 20log(2)

Look at the last example there. We just simplified the gigantic 2²⁰ to a reasonable looking 20log(2), which is the formula to figure out the number of digits it has. In other words, the number of digits in 2²⁰ is just 20log(2) rounded to the nearest integer. Plugging this into a calculator tells you that the log is 6.0206, meaning that there are seven digits in the number. If you multiply it out, you will find that 2²⁰ = 1048576, which does indeed have seven digits. 

So whenever a type of problem pops up with a power of this sort, try to determine how many digits it is. Chances are you will gain a much better understanding of the statistic when you perform this quick calculation.

Logarithmic Proofs and Identities

I have talked about logarithms quite a bit on this blog, but they were always being applied to something else, whether it be Benford's Law, the Law of 72, or some other practical use. This week, I would like to show that logarithms are very much a part of pure mathematics as well. Of course they are in mathematical equations just as much as exponents and radicals, but they have some pretty cool features of their own.

First off, let me review what a logarithm is. I have explained it before, but once you understand the notation, you shouldn't need to have done Precalculus to understand this post. They are very easy to understand.

For instance, we know that 10² is 100.

10² = 100

If we take the logarithm of both sides, we are essentially bringing the two out of the exponent. Rather than doing an operation on the 10, we do an operation on the 100 to determine what the exponent is.

log(10²) = log(100)
2 = log(100)

In most situations, it is clear what type of logarithm you are using, especially in this one because ten is a common logarithm to use. However, many people will write a subscript to clarify. For instance:

log₁₀(100) = 2

Logarithms become extremely useful when you need to solve an algebraic equation where the variable is in the exponent. For example:

2^x = 64

As you know, algebra is about doing the inverse operation. If there is addition going on, you subtract. If there is multiplication going on, you divide. Similarly, if there is exponentiation going on, you use a logarithm. In this instance, it would be taking the log₂ of both sides.

c2^x) = log₂(64)
x = log₂(64)
x = 6

In practice, there are three bases that are extremely popular to use in a logarithm. We just used two of them: log₁₀ and log₂, which are also known as the common logarithm and the binary logarithm. The third one is log_e (using the number e that is described here), which is called the natural logarithm, or the natural log. This one is found on most calculators, usually next to the common logarithm.

Though logarithms are not a part of most people's day-to-day life, they do have lots of practical applications. The common logarithm is the basis of the pH system which describes the acidity of water. The natural log is a huge aspect of finance and compound interest (as we saw with the Law of 72). And of course, they are all over nature.

Let's look at an identity of logarithms. Take the following problem:

log₆(24) + log₆(9) =

If you just used a calculator to do this, you would get:

log₆(24) + log₆(9)
1.773705614 + 1.226294386
3

That's odd. Two random numbers happened to have logarithms that summed to three. Let's look closer at this and see if we can figure out why. What number could you find the log₆ of and get 3?

log₆(n) = 3

First of all, let me point out that we just asked an algebraic question. We suddenly got curious about why something happened, so we asked a "what" question, which calls for an unknown quantity, which later becomes an algebraic variable. So when algebra seems like a drag, remember that it is all techniques for answering that "what" question. And "what" is a question asked in all branches of mathematics, science, and engineering.

Anyways, for this equation, we would want to do the inverse operation. We turn both sides into the exponent, and create a base of 6. This gives:

6^log₆(n) = 6³

The left hand side cancels, leaving just n. The right hand side is six cubed, which is 216. So, we end up with:

n = 216

So, this means that the log₆ of 216 makes you end up with 3, or the sum of log₆(24) and log₆(9).

log₆(24) + log₆(9) = log₆(216)

What is the relationship between these three numbers? Well, it shouldn't take to long to determine that 24 x 9 = 216, or:

log₆(24) + log₆(9) = log₆(24 • 9)

In other words, the sum of the logarithms is the logarithm of the product. Wow! That's pretty cool! Is that always the case? Well, let's try to prove that it is for all logarithms.

Let's set a few terms equal to each other and see what happens. Since there are logarithms, we will need a lot of variables.

x = log_a(p)
y = log_a(q)

In other words:

p = a^x
q = a^y

Let's multiply those two equations together. Since they are both equal, multiplying the terms on each side by each other won't make a difference.

p • q = a^x • a^y
pq = a^x+y

The right hand side was simplified using the Law of Exponents, which is explained very well here.

Now, we must take the logarithm of both sides, or specifically, the log_a of both sides.

log_a(pq) = log_a(a^x+y)
log_a(pq) = x + y

But what were x and y? We defined them in terms of a, p, and q earlier. So, let's substitute those values in and see what we get.

log_a(pq) = log_a(p) + log_a(q)

And this creates the identity that we were trying to prove: the sum of the logarithms is the logarithm of the product, and thus, completes our proof. There are other logarithmic identities like this one, but I will save that for another post.

A Quick Way to Check Your Work

In school, the teacher is always on top of you for checking your work. When you do a subtraction problem, solve the reversed addition problem and make sure it is right, when you do an algebra problem, make sure you plug your solution back into the original equation. These are all things that are drilled into our heads, but never quite executed.

I did post a year and a half ago about checking your work in algebra problems: plugging the answer into the original equation (click here to see how to do that). But there is also a shortcut for checking work on plain arithmetic problems as well.

Let's take the problem 138 + 253. I would have went smaller, but the method will be easier to demonstrate with larger numbers.

   138
+ 253

If we add that up normally, we would get:


   138
+ 253
   391

How do we know if that is correct? Well, we do something called mod sums. What that means is we add up the digits in the number, and then add up the digits in this sum, and keep going until we find a single digit number. This is called the number's mod sum or digital root.

So, what is the mod sum of 138? Well, we add up the digits.

1 + 3 + 8 = 12

1 + 2 = 3

So, the mod sum or digital root of 138 is 3. Let's find it for 253.

2 + 5 + 3 = 10

1 + 0 = 1

The mod sum of 253 is therefore 1. Let's find the mod sum of the total and see if you notice the pattern.

3 + 9 + 1 = 13

1 + 3 = 4

So, the two addends have mod sums of 3 and 1. The sum has a mod sum of 4. What is the pattern? That's right, the mod sum of the answer is the sum of the mod sums of the addends. What about a subtraction problem?

  924
- 643

The answer to this problem is 281. But how do we confirm it?

The mod sum of 924 is 6 (9+2+4=15 and 1+5=6) and the mod sum of 643 is 4 (6+4+3=13 and 1+3=4). So, the mod sum of the difference must be the difference of the two mod sums. The mod sum of 281 is 2 (2+8+1=11 and 1+1=2), which is the difference of 6 and 4. So, the answer was correct.

What about a multiplication problem? Say 71 x 55. If you do the math, you will find that the answer is 3905. But let's check it with mod sums.

Mod Sum of 71 = 8
Mod Sum of 55 = 1
Mod Sum of 3905 = 8

8 x 1 = 8

So it is correct. There are some glitches in the technique, but this is the basis of it. You might run into scenarios that I didn't quite explain how to deal with, but feel free to comment. I will be happy to respond with some more specific pointers. Have fun actually checking your work now!

Carl Friedrich Gauss: The Child Prodigy

One of the most famous mathematicians of all time is the German mathematician Carl Friedrich Gauss (1777 - 1855). Gauss was one of the leading number theorists of all time, as well as a contributor to algebra, statistics, geometry, analysis, and applied mathematics.

There is a piece of mathematical folklore (which may or may not be 100% accurate) that involved a child Gauss. It is a great story, highlights a great point, and shows the intelligence of a great mathematician.

A fifth grade teacher is teaching a class, and started to get frustrated with the students. So, in an attempt to punish them, she demanded that they add up all of the numbers from 1 to 100. This is a daunting task for the average person. She expected to have the students start working on the problem, and she could leave and take a break.

As she was about to walk out the door, the young Gauss raised his hand and declared that the answer is 5050. The teacher was stunned. After checking his work, they found that 5050 was the correct answer.

How did he do it? Well, he visualized a horizontal line with all 100 numbers:

1   2   3   4   5   ...   96   97   98   99   100

And then he took the second half of that line (51 - 100) and flipped it around underneath to look like so:

    1     2     3     4     5   ...   46   47   48   49   50

100   99   98   97   96   ...   55   54   53   52   51

Each of these vertical columns is its own addition problem. And in all fifty columns, the sum is 101. So, the sum of the numbers from one to one-hundred is the same as fifty 101's, or 50 x 101. Since 50 ends in a zero, it is a pretty quick computation: 50 x 101 = 5050. And there is the answer.

I think this is a great story when it comes to historical mathematicians, regardless of how true it is. Gauss did go on to study triangular numbers, which are the sums of consecutive integers up to a point. And since triangular numbers are absolutely fascinating, this story is a great way to begin an endeavor in that topic.

Don't Borrow Tens Ever Again in a Subtraction Problem!

When the traditional method of subtraction is taught in school, it can often be quite confusing, especially when the number you are subtracting from has lots of zeros. It is a pain to go borrow the tens from other numbers, especially when they are ones or zeros. But, there is an easier way to approach this that isn't taught in class.

In geometry, the word "complimentary" is used to describe the relationship between two angles that sum to 90°. For instance, 36 and 54 are complimentary angles. Similarly, the compliment of 36 would be 54 when talking about geometry.

A geometric application of complimentary angles (and a funny picture)

With arithmetic on the other hand, a number's compliment is its difference from the power of ten above it. So, the compliment of 36 would be 64 (100 - 36 = 64), or the compliment of 473 would be 527 (1000 - 473 = 527). This is more practical for solving arithmetic problems than the distance from 90.

One-hundred or one-thousand are very difficult numbers to subtract from. These are ones where you need to keep turning zeros into nines until you reach that first digit, and then figure out the problem from there. This can be quite the pain. But, there is a shortcut for these examples.

All you have to do is subtract the last digit of the number from ten. This is the last digit of the compliment. Then, subtract all of the rest of the digits from nine, and place them in their corresponding place value. For instance, let's find the compliment of 36.

10 - 6 = 4
9 - 3 = 6

100 - 36 = 64

Pretty easy, right? Let's try it with 473.

10 - 3 = 7
9 - 7 = 2
9 - 4 = 5

1000 - 473 = 527

This can easily be extended to numbers in the thousands, millions, billions, and more as long as you can keep track of the digits. For example, 5647823 would have a compliment of:

10 - 3 = 7
9 - 2 = 7
9 - 8 = 1
9 - 7 = 2
9 - 4 = 5
9 - 6 = 3
9 - 5 = 4

10000000 - 5647823 = 4352177

Couldn't be easier! This method is basically just remembering the fact that no matter what the situation, you will have borrowed a ten from each zero, making them all nines except for the very last one which will remain a ten. Try a few subtraction problems yourself and you will see why it works.

This can be applied to geometry in a way as well. To subtract quickly from ninety, you still subtract the last digit from ten, and the first one from eight (one less than the nine). So, the compliment of 29° is:

10 - 9 = 1
8 - 2 = 6

90° - 29° = 61°

For supplements of angles (the angle's difference from 180), you can do a similar technique as well. Subtract the last digit from ten and the first from seventeen. For instance, the supplement of 48° is:

10 - 8 = 2
17 - 4 = 13

180° - 48° = 132°

In fact, any number ending in zero(s) can be subtracted from just by altering this method. I have found this extremely helpful when performing mental math (three-digit and four-digit squaring requires you to quickly identify how far you are from the nearest hundred/thousand, which often needs compliments). It is also very practical. When you give the cashier a hundred dollar bill, they are usually impressed when you tell them the change before they have time to punch the bill into the register. I'd recommend practicing this technique because it is useful and also quite a bit of fun.

Multiplication of Fractions Proof

Let me show you a pretty cool proof. We will prove that:

1/2 • 3/4 • 5/6 • 7/8 • 9/10 • ... • 99/100 < 1/10

With most proofs, it is easy to see if the conjecture is correct on face, but the actual proof takes more work. In this example, it is extremely difficult to tell if 1/10 would be less or greater than the product.

Let's start by denoting the left-hand side with A.

A = 1/2 • 3/4 • 5/6 • 7/8 • 9/10 • ... • 99/100

We will also denote the sort of opposite of A with B.

B = 2/3 • 4/5 • 6/7 • 8/9 • 10/11 • ... • 98/99

You will notice that every term in the B product has a corresponding term in the A product. 1/2 and 2/3 are both the first term, 3/4 and 4/5 are both the second term, and so on. You will also notice that every term in the B fraction is greater than every term in the A fraction. So, we can conclude that B is greater than A, or A is less than B.

A < B

Let's multiply both sides of that inequality by A.

A • A < A • B

A² < AB

The product of A and B is something that we can figure out pretty easily. Since the numerators of A are the denominators of B and vice versa, everything will cancel out. All that will be left is 1/100, since 1 and 100 were not in the B product. So, AB = 1/100. Substitute 1/100 in for AB to get:

A² < AB
A² < 1/100

If we solve this inequality for A, we will find that A < 1/10.

A² < 1/100
√(A²) < √(1/100)
A < 1/10

But what is A equal to? At the beginning, we denoted it to be equal to the left-hand side of the hypothesis inequality. Substitute that in for A, and we get:

1/2 • 3/4 • 5/6 • 7/8 • 9/10 • ... • 99/100 < 1/10

And there is our proof. These types of proofs don't really innovate mathematics, but I do think they can be fun.

Multiplication Made Visual

In school, multiplication is taught by giving students a specific method to use (which they call the "traditional" method), and you have to solve problems with it. By only providing one of over a dozen multiplication methods, this eliminates the creativity, and quite possibly the fun, in multiplication. I'm not going to dwell on these issues; you can read my Capstone Research Paper for that.

I have discussed a few other methods of multiplication on this blog, one of which is the Criss-Cross Method. Again, I won't explain it here, but this method does lay the foundation for the new method I am going to explain today.

Rather than taking you through the steps, I will post a video tutorial on how to do it (those are usually more fun anyways). If you are wondering why it works, I encourage you to refer back to the Criss-Cross Method and notice the similarities. Try a few problems with the Criss-Cross Method, and then this method, and see how the numbers that appear throughout the problem seem to be identical.

When you watch this video, I would strongly suggest doing the examples along with the video. It can be a very useful technique for problems whose digits are of a reasonable size.

I think this is a really cool method of multiplication, and even easier to teach than the traditional method. Play around with it, and you will likely end up using it in the future.

For some more intellectual readers, it might be fun to try multiplying numbers with bigger digits using the quinary or senary number systems (base five or base six). This would make less dots in each section. I think it might be a fun exercise.

Answers to June Problem of the Week:

*For the answers to the creation of a line of best fit, it will be written as an inequality. There is no way to get the exact answer by hand, so just make sure your answer is in the interval that is written.

Easy:
a = 7¹/₃
p = 14

m = ¹¹⁸⁹⁸⁸/₆₉₂₅
n = ³⁹⁴/₂₇₇
y = 15.76
z = 12.08
g = 8
b = 112
q = 44
x = 33

Medium:
h = 105
p = 270
t = 19
n = 92925
u = 92.925
q = 135
a = 2165.2875
d = 21.652875
g = 15
j = 46.4625
k = 23.23125
r = 27
s = 45
2 < m < 3.5
45 < b < 50
175 < x < 275

Hard:
s = 107
t = 365
p = 905
q = 4096
q₀ = 4096
q₁ = 2048
q₂ = 1024
q₃ = 512
q₄ = 256
q₅ = 128
q₆ = 64
q₇ = 32
q₈ = 16
q₉ = 8
q₁₀ = 4
q₁₁ = 2
q₁₂ = 1
q₁₃ = ¹/₂
q₁₄ = ¹/₄
q₁₅ = ¹/₈
x₁ = -452.5
x₂ = -2
x₃ = 4
y = 0
a = 226.25
b = -224.25
c = -454.5
t = 2635773.77
m = 34
x = 79.53%

The Magic of Algebra

Think of a number from one to ten.
Multiply that number by two.
Add thirteen.
Subtract seven.
Divide by two.
Subtract your original number.
You are thinking of the number three.

This is a classic mind reading trick that many people will learn to do when they are little. The premise is simple; the answer is always three.

Why does this work? As a young child, you probably think back to the arithmetic skills you learned in school and realized that you can't prove this trick because you don't know what number they'll choose.

This is a good time to note that when you learned this trick, the first thing you thought was "why?" Many people see proofs as a tedious process that could be avoided just by accepting things to be true. However, that why spark that went off when you learned the trick is where proofs can be applied. I use proofs so often in my posts primarily for that reason: to foster that why spark.

Anyways, this situation is where you need to define that unknown quantity of the number the spectator is thinking of. How do you define an unknown quantity? Well, this is when you pull out a little tool called algebra. You might not expect to apply algebra to a practical situation, but this is when you can. With algebra, we can stop saying "the number that they are thinking of" and start saying "n."

n

Now, the first step was to multiply the number by 2. n times 2 is pretty simple; it is 2n.

n • 2
2n

Next, we had to add thirteen. That can't be simplified much.

2n + 13

Now, we subtract seven. 13 - 7 is 6, so we can simplify it to 2n + 6.

2n + 13 - 7
2n + 6

Now, we have to divide by two. 2 and 6 both have a factor of two in it, so we will not need any fractions.

(2n + 6)/2
2(n + 3)/2
n + 3

The last step was to subtract the original number. But, we already defined that number to be n, so we just subtract n.

n + 3 - n
3

As you see, the n's cancelled, and you ended up with three. Basically, you undid all of the previous operations to end up with one number.

I think that this is a perfect way to show young children magic, but also introduce them to concepts like algebra and proofs by lighting that spark in such a fun and interesting way.

History of Math: Leonhard Euler

Today, I am giving a talk about Mending Mathematics Education in America at TEDxBushnellPark in Hartford, Connecticut. I did a post on this issue (click here to see it), as well as a huge school project on the issue (click here to see the research paper).

In the talk, I will be bringing up a famous math problem that originated with Leonhard Euler. I won't post about that problem today, but I will give a little story from the life of Euler.

Leonhard Euler was born on April 15, 1707 in Basel, Switzerland. He chose the path of a mathematician (if he didn't, I probably wouldn't be talking about him right now), and ended up as one of the most prolific mathematics writers of all time.

Some of his biggest contributions include standardizing the notation for the number e and π, coming up with Euler paths and circuits, and revamping most of the branches of mathematics that were known in his time. He was said to be able to entertain a child, scratch a cat, and calculate math problems simultaneously.

On this blog, I have mentioned sequences a lot. For instance, the triangular numbers are a sequence. The negative-first powers are a sequence. The powers of 1/2 are a sequence. Let's look at these three sequences:

1, 3, 6, 10, 15, 21, 28, 36, 45, 55...
1, 1/2, 1/3, 1/4, 1/5, 1/6, 1/7, 1/8, 1/9, 1/10...
1, 1/2, 1/4, 1/8, 1/16, 1/32, 1/64, 1/128...

We haven't done it before, but a common thing to do is add up all of the numbers in these sequences. In some sequences, this seems odd to be doing, but in other ones, it seems completely understandable. Let's try it with those three.

1 + 3 + 6 + 10 + 15 + 21 + 28 + 36 + 45 + 55 +...

The individual numbers themselves are going towards infinity, so the sum of them is probably infinity as well. Generally, if the numbers in a sequence are growing, then the sum will be infinite. This type of sequence is called a divergent sequence, and questions like their sum are generally not asked.

What about the next series. This one is a little harder to tell, but if we group some things together, it will be easier to see.

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

Compare this to the following sequence, which is clearly less than this one.

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

Since all of the terms in the first sequence seem to be bigger than the second one, then the first one has to sum to a greater total. But, the second sequence simplifies to 1 + 1/2 + 1/2 + 1/2 +..., which will end up as infinity. So, a sequence that has a bigger sum must also be infinite as well. This makes that a diverging sequence.

What about the third one? Well, combining terms together won't do much, but let's try figuring out what the sum is by setting the sequence equal to x.

x = 1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 +...

Now, what if you were to multiply that by two? You would get:

2x = 2(1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 +...)
2x = 2 + 1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 +...
2x = 2 + (1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 +...)
2x = 2 + x
x = 2

So, the sum of this sequence is not infinity, but rather the natural number two. Since the sequence does not diverge to infinity, it is called a convergent sequence. These are the type that are generally summed to their total, since that number can then be applied.

Anyways, some of Euler's students were analyzing a much more complicated convergent sequence to figure out its sum. They had been adding numbers, and after they got to the seventeenth number in the sequence, they found their answers to be different at the fiftieth decimal place. Since they didn't want to go back through that tedious process, they argued and argued over who got it right.

Eventually, they went to their professor, Euler, to ask him for assistance. After asking for the problem, he mentally calculated out the answer to the fiftieth decimal, and was able to determine who was correct.

I found that story kind of funny, but also really impressive. To be able to calculate that difficult of a problem to that high of a degree shows some real mathematical talent.

Math in the News: Facebook Math Problem

A few weeks ago, I suddenly opened up my Facebook page to see that I was tagged by several people in a post about this article. Once I saw it, I immediately wanted to discuss it on Cool Math Stuff.

First, try to solve this math problem:

6 ÷ 2(1 + 2) = 

Even though it seems simple, there is lots of debate about the answer on social networking sites. Click here to view the article.

When I first saw the problem, I thought the answer was undoubtably 1. When I did the problem, I had multiplied by two while I was simplifying the parentheses. So, I assumed that people who got 9 as the answer were forgetting the order of operations.

Before I talk about the article, I was excited to see that this was a debate between people. Any time where the general public is talking about mathematics, especially in a argumentative way, is a huge accomplishment for society.

Reading the article, I was intrigued to find myself in the minority. I did not think of the parentheses as a substitute for the multiplication symbol, but a quick way to write two times the quantity one plus two.

What if the problem was written like this?

6 ÷ 2 • (1 + 2) =

If the problem were written like that, I would have gotten 9 as my answer. Since the 2 was separated from the parentheses, "Please Excuse My Dear Aunt Sally" would put the division first and then the multiplication.

However, without the dot there (more formally known as an interpunct), I do feel that the answer should be one. It seems like the most practical perspective on the problem.

Take an expression like 2a. It is implied that the 2 and the a are a single quantity. If you were using it in an actual problem, it is far more likely that you would be dividing, say 8, by the whole quantity 2a than 8 by 2, and then multiplying this by a.

Similarly, let's say you had 2(a + 3). Again, there would be no doubt that you should distribute the 2. This problem would then become 2a + 6, which seems like the correct thing.

If you ended up with 1 ÷ 2(a + 3), there would be no practical case where the one were to be divided by the two first.

Also, I think that the order of operations was put into place to make algebra and geometry more structured. For instance, say you had rhombus where the diagonals cross, or bisect, at a (3x³ + 9)° angle, and you had to find the measure of the sides which were of length 25 - 2x². As you probably know, a rhombus is a quadrilateral with all four sides equal (a square has all four sides and all four angles equal).

There is a theorem that states that the intersection of the diagonals of a rhombus is always equal to 90°. I think this is pretty cool on its own, to know this angle all the time. So, we can assume that 3x³ + 9 must equal 90°.

3x³ + 9 = 90
3x³ = 81
x³ = 27
x = 3

We must then plug this into the expression equal to the side measure to find our answer. First, let's look at the expression.

25 - 2x²

What are we being asked to do? Since two is a coefficient for x², we would square the x first, then multiply by 2, and then subtract that from 25. This would be agreed upon by all algebra teachers. What happens if we plug the 3 back in?

25 - 2(3)²

If there was no order of operations, we would probably do this problem from left to right. This would give us:


25 - 2(3)²

23(3)²

69²

4761

You can probably already tell that 4761 is way bigger than this side length would be intended to be. Even if you didn't make this estimate, this attempt at the simplification was nowhere close to the agreed order. By implementing the order of operations, there is no debate that the order should be PEMDAS. This clarifies lots of algebra when you are substituting terms into equations.

What if the expression were 108 ÷ 2x²? I still think most of us would agree that the 2x² is meant to be its own term, and would therefore be divided as a quantity into the 108.

But after substitution, we get the following:

108 ÷ 2(3)²

108 ÷ 2(9)

If we did the 108 ÷ 2 first, we would end up with the wrong answer. Even though the surface definition of PEMDAS would ask for this to be done first, it is not a practical approach to the problem.

However, I did see that the SAT or ACT would expect students to receive 9 as an answer. Since there is clearly not a correct perspective to have, I would encourage you to comment what you thought the answer to be and why. Just like with Pi vs Tau, this is a post that is actually a lot of fun to debate about.

Benford's Law

Take any random quantity. Maybe a population of a city, a mass of a planet, a distance from a star, or an amount of twitter followers. Now, take the first digit of this number. What are the odds that it will start with a 1, or a 2, and so on?

One would think that it would be a one in nine chance for each digit. If you are randomly selecting this number, the first digit is just a random number selection between one and nine.

However, this reasoning does not work. Try finding these quantities yourself. Or, just go to http://testingbenfordslaw.com. If you check all of the different areas that are possible, you will see that the smaller the digit, the more commonly it appears.

Here are the approximate odds for each digit:

1: 30.1%
2: 17.6%
3: 12.5%
4: 9.7%
5: 7.9%
6: 6.7%
7: 5.8%
8: 5.1%
9: 4.6%

Since this is a math blog, something we would want to do is find a pattern between these numbers. This could probably be done on a graphing calculator, using similar techniques to the post on data analysis.

It is clear that a line of best fit would not be the solution. If we connected the points with a curved graph, it would look like this:

You can see that this graph gets really close to the y-axis, but it does not seem to touch it. Similarly, it gets really close to the x-axis, but doesn't touch it. Thus, the x and y-axes would be called asymptotes of this graph.

But, an asymptote could be found in a rational function, radical function, exponential function, logarithmic function, hyperbolic function, trigonometric function... So, it is hard to define this graph solely based on the presence of asymptotes.

The function that does work with this graph is a logarithmic one. A logarithm is basically the opposite of an exponent. For instance:

5² = 25
log₅(25) = 2

Just knowing that the equation is logarithmic doesn't seem to narrow it down a lot, because there are so many different types of logarithms. However, there are three types which are seen the most frequently. In fact, most scientific calculators contain just these three types.

Logarithm Type	Base	Simple Notation	Standardized Notation	Applications
Natural Logarithm	e (the irrational number ≈ 2.71828)	loge(x)	ln(x)	Calculus, Statistics, Physics, Chemistry, Economics
Common Logarithm	10	log10(x)	lg(x)	Algebra, Engineering, Geology, Spectroscopy, Music
Binary Logarithm	2	log2(x)	lb(x)	Discrete Mathematics, Computer Science, Information Theory

If we confine the possible equations to one of these, it narrows it down a lot. And this assumption is correct. The equation that the graph fits is:

y = lg(1 + 1/x)

(Remember that x is the starting digit and y is the percentage for that digit)

I was pretty surprised about this equation. But, all of the percentages listed above are the outcome of this equation.

Benford's Law can be used in biology, accounting, law, economics, etc. However, a more fun way to use it is to turn it into a game.

Tell someone that you will get the numbers 1, 2, and 3 and they will get the numbers 5, 6, 7, 8, and 9 (and nobody gets 4). You then have them come up with random quantities that they wouldn't know, and you look up the number (you can use Google or WolframAlpha for this). Every time it is a 1, 2, or 3, you win a point and every time it is a 5, 6, 7, 8, 9, they get a point. They think they have around a 2:1 advantage over you, but you are really the one with the 2:1 advantage.

Click here to see this game played on the show Scam School.

Handshake Problem

Let me start by giving a little math problem. n people are in a room, and each one has to shake every other person's hand. How many handshakes will it take for all n people to have shaken every other hand?

Let's first try a few numbers. For two people, it would obviously just take one handshake. For three, it would take three handshakes. For four, it would take six handshakes. For five, it would take ten handshakes.

Do you see the pattern? 1, 3, 6, and 10 are the triangular numbers. In fact, this pattern always continues. 

If you look at it logically, you will see why. Person 1 has a number of hands to shake. Person 2 would have to shake all of the hands except for person 1's (it was already counted). Each person has to shake one less than the one before until there is just one left.

Adding up these handshakes will be a sum of the first n natural numbers, which is the definition of triangular numbers. 

I found it cool that a famous number sequence could be applied to a practical problem like this. 

Turning Numbers Rational

Today, I wanted to do something that I haven't done in a while. I wanted to prove something, but with no patterns involved. Just take a question and prove the answer.

What I want to prove is that an irrational number to an irrational power can be rational. Like, could you raise π to a power and get a whole number, or a simple fraction?

π is a very hard number to work with since there is not a clean way of deriving it (click here to see how you can derive it), so let's take a different irrational number and use that for the example. Say, the square root of 2. We proved it irrational in November of 2011 (click here for that), so we can use that for this example.

Let's take (√2)^(√2). What does that equal?

Most people, myself included, would just say that they don't know. That is a good answer. Let's stick to two possibilities.

1. It is a rational number (it can be written as a fraction, so it is a terminating or repeating decimal)
2. It is an irrational number (it cannot be written as a fraction)

If it is a rational number, we are done. We proved that an irrational number to an irrational power is rational. That would be easy!

What do we do if it's irrational? Since we know it's irrational, we can use it as the irrational number being raised to the irrational power. Let's just see what happens if we use [(√2)^(√2)] as the base and (√2) as the exponent.

[(√2)^(√2)]^(√2)

The law of exponents says that (a^b)^c = a^bc, so we can use that to our advantage.

[(√2)^(√2)]^(√2)
(√2)^(√2)(√2)
(√2)^2
2

So, we ended up with a rational number. This means that if it were irrational, we have proven the original statement true as well. For either of the two possibilities, we have a proof.

How to Divide Any Number by 91

Since this is the last post of 2012, I thought I'd finish the year with a mental math technique. Before I do that though, I want to show you a pattern that is closely related.

Take any three digit number. For this, I will use 123.

123

Multiply that by 13. You should get 1599.

Now, multiply that by 11. You should get 17589. You might have done that in your head using the multiplication by eleven trick I taught last October.

Now, multiply that by 7. This will give you 123123. And if you'll notice, we started with the number 123 and finished with two 123s.

Why did this happen? It is very simple, and just relies on the fact that 13 x 11 x 7 is 1001. So, by multiplying by these seemingly random numbers is really multiplying by 1001. And any number times 1001 is just itself repeated twice.

Division by 1001 results in a similar answer. For instance, 123 ÷ 1001 = 0.122877122877...

If you notice, the first three digits are 122, which is just 123 - 1. The next three are 877, and 122 + 877 = 999.

This pattern continues as well. I am not sure how to prove that, but please comment if you do know.

So, to divide by 1001, you just subtract one for the first three digits, and subtract the first three digits from 999 for the next three digits.

However, telling someone you can divide any number by 1001 doesn't sound that impressive. Since 1001 is right next to 1000, people will get very suspicious.

That is why I showed you the multiplication pattern. Dividing by 1001 is basically dividing by 13, then 7, then 11.

123 ÷ 1001 = 123 ÷ (13 x 7 x 11)

So, instead of going all the way to 1001, let's just get part way there by dividing by the 13 x 7, which is 91. We will use 33 as the number we are dividing by, or the dividend.

33
91

Our goal is to make it a number divided by 1001, since we know how to do that. That means that the first step is multiplying the 91 by eleven. But, if we multiply the denominator by eleven, we must also multiply the numerator by eleven.

33 x 11
91 x 11

We know 91 x 11 is 1001, so we have the problem we are looking for. What is 33 x 11 though? If you go to the multiplication by eleven post, you will see that it is just 363 (add the 3 + 3, and stick it in the middle). This gives us the problem 363 ÷ 1001.

363 - 1 is 362 and 999 - 362 is 637, so the answer is 0.362637362637...

So, to divide by 91, you just multiply the number by 11, subtract one, and subtract that from 999 to get the answer.

Though it is a lot tougher, you can also divide by 77 (11 x 7) and 143 (13 x 11) with the same principle. You are just multiplying the top number by 13 or 7 instead of the easy 11.

The 1001 pattern is really cool, and something you can turn into a really cool trick. The multiplication by eleven is also a really cool effect. But putting them together into the division by 91 is a really impressive feat of mental math that will fool all of your friends.

Triangular Day: How to Square ANY Number

Today is another triangular day! It is the fifteenth of December and 15 is the fifth triangular number.

Two weeks ago, I talked about how the triangular numbers are one of the many figurative families, which are groups of numbers whose elements form the corresponding equilateral polygon. The triangular numbers are the first of these families, because its elements form an equilateral triangle.

The next of these families were the square numbers, and they are the same square numbers you are thinking of. You can easily find a square number by doing n^2, where you square n to get the nth square number.

Today, I want to continue discussing square numbers, but in more of a fun way. But before we get to the fun part, I want to do a little algebra.

We may remember the binomial theorem from algebra. It said that:

(a + b)(a + b) = a^2 + 2ab + b^2

You also might remember the reverse of this.

(a - b)(a - b) = a^2 - 2ab + b^2

But what if we do one plus and one minus?

(a + b)(a - b) =

It isn't very obvious with the simplification of the other two examples. So, let's just do it out.

(a + b)(a - b)
a^2 - ab + ab - b^2
a^2 - b^2

(a + b)(a - b) = a^2 - b^2

There is the simplification. It is commonly called the difference of two squares, and is an easy way to factor quadratic equations. For example, if you had:

f(x) = 4x^2 - 25

You could factor it easily into:

f(x) = (2x + 5)(2x - 5)

This topic comes up in probably every Algebra II class, but nobody ever notices just a little alteration you can make to it.

Let me move to the subject I want to talk about today, which is squaring numbers. You may know that I perform a mental math stage show called Mathemagics, where I do many feats of mental math. One of the most popular of them is the one where I ask an audience member for a number and I square it in my head.

People always jump to the conclusion that it is either super easy (like a hidden calculator gimmick or an unexplainable gift) or super hard (like a complicated formula that no one would ever be able to learn without extensive training). The super easy explanations are completely inaccurate, and the super hard explanations are also fairly incorrect. They are right that I practiced a formula, but it really isn't that complicated. In fact, you are indirectly taught it in your Algebra II class.

Let's try an example, like 18^2. 18 isn't too bad to multiply by, but what number close to 18 is easier?

20. So, we will go up two to twenty and that means we have to go down two to sixteen.

20
 |
18
 |
16

So, the first thing we do is 20 x 16. That may sound tough, but remember that it is only 2 x 16 with a zero tacked onto the end.


20
 |   \
18   320
 |   /
16

We are almost done. All we have to do is add to that the square of whatever number we went up and down. 2^2 is four, so we do 320 + 4 to get 324. And there is the answer.

Let's try a tougher one, say 67^2. How about you try it in your head and see if you figure it out.

Okay, this is hard for a second example. 67 is close to 70, so we go up three to 70 and down three to 64.


70
 |
67
 |
64

Now, 70 x 64 doesn't seem any harder than 7 x 64, but you can pull it off. Rather than trying what you learned in school with right to left multiplication, you have to switch to left to right multiplication. This sounds weird, but it is the correct way to approach it mentally.

7 x 60 is 420. 7 x 4 is 28. 420 + 28, left to right, is 448. Tack on a zero to get 4480.


70
 |   \
67   4480
 |   /
64

Now what do we do? We add the square of what we went up and down, namely the square of three. 3^2 = 9, so we get:


70
 |   \
67   4480
 |   /  +   9
64   4489

And there's the answer. For three digit numbers, you do the same general thing. Try 381^2. 381 is 19 away from 400, so we go up to 400 and down to 362.

400
  |
381
  |
362

400 x 362, left to right is 144800, and now we need to figure out 19^2.


400
  |    \
381   144800
  |    /
362

19 is one away from 20, so we go up to 20 and down to 18. 20 x 18 is 360 plus one squared is 361.


400
  |    \
381   144800
  |    /    + 361
362   145161

And there is the answer. With just a couple weeks of practice, this will become second nature.

But why does this work? If you'll remember from our brief Algebra II review, it was right in our face.

a^2 - b^2 = (a + b)(a - b)

We just have one tiny alteration to make. Let's add b^2 to both sides.

a^2 = (a + b)(a - b) + b^2

And there you go. The number we are squaring is a, and the number that we go up and down is b.

Lots of the stuff I put here is stuff that I say you can easily implement into school curricula, but this is probably the easiest thing to put in. All the teacher has to do is mention that if you add b^2 to both sides, you have yourself an easy squaring formula and maybe demonstrate it.

This was very abnormal compared to the other triangular days. But if you read the post of any triangular day, I would suggest reading this one.

If you'd like to see me actually squaring numbers, you can find video footage on my website ethanmath.com, or my greatest hits video which I have posted on the blog.

Criss-Cross Method: Multiplication Made Easy

In school, you are taught pretty early on how to do multiplication on paper. For instance:

  47
x  8

That would be tough to pull off in your head. However, in school, you learn that you can simply do it in two steps.

1. 7 x 8 is 56. Write the 6 and carry the five.
  5

  47
x  8

    6

2. 4 x 8 is 32. Add the five to that and you get 37.

  5

  47
x  8
376

And there you go. This works great for small problems like 2x1, 3x1, 4x1, 5x1, or even ones like 2x2 or 3x2. However, what about something like a 4x4. Your paper would look like:

            8922
x          6137

          62454

        267660

        892200

+  53532000

    54754314

This method is basically the distributive property, where you are multiplying each number by every other number. For instance, a 3x3 algebraically would look like:

Let x = the first digit of the first factor

Let y = the second digit of the first factor

Let z = the third digit of the first factor

Let a = the first digit of the second factor

Let b = the second digit of the second factor

Let c = the third digit of the second factor

Let P = the product of the two factors

P = (z + 10y + 100x)(c + 10b + 100a)

Using this traditional multiplication method, this product then looks like:

P = c(z + 10y + 100x) + 10b(z + 10y + 100x) + 100a(z + 10y + 100x)

This looks pretty complicated. What this method then does is takes it one step further by distributing the a, b, and c through their parentheses. The one thing that it leaves factored out is the powers of ten that are coefficients of the a, b, and c. You don't multiply through the 10 and 100 that are in front of the b and the a.

P = [(c)(z) + (c)(10y) + (c)(100x)] + 10[(b)(z) + (b)(10y) + (b)(100x)] + 100[(a)(z) + (a)(10y) + a(100x)]

You might be thinking that you don't do that much work in your head. I mean, you've been doing this since third grade! But think through what you are doing. How are you creating the terms to add up?

The first term is created by multiplying the last digit of the second factor, or c, by each digit of the first factor. There is a multiplication method called partial products which would multiply the c by the 10y and 100x, but the traditional method uses the power of zero rule to just write the (c)(10y) to the left of the (c)(z), and the (c)(100x) to the left of that.

The second term is created similarly by multiplying the second digit of the second factor, which we denoted as b, by each digit of the first factor. However, we must remember to put a zero down first, which was something constantly driven into our heads by our math teachers. This is taking into account the ten in front of the brackets. The power of zero rule states that when multiplying by ten, we can just tack a zero onto the end of the other factor. So, we tack a zero onto the end of this number that was found by multiplying b by every digit of the first factor.

The third term is also created by multiplying the first digit of the second factor, which we called a, by each digit of the first factor. Since we have a 100 outside of the brackets, we must tack two zeros onto the end of this number.

Finally, we add up these terms, which is exactly what the equation does above to get the product.

Is this really how we want to get to our product? Can't we combine the terms in a little bit of an easier way?

Let's try it. Let's fully distribute every single term through. We will get:

P = 1cz + 10cy + 100cx + 10bz + 100by + 1000bx + 100az + 1000ay + 10000ax

How about we treat the coefficients as if they are variables. In that case, we have lots of like terms to add.

P = 1(cz) + 10(cy + bz) + 100(cx + by + az) + 1000(bx + ay) + 10000(ax)

That looks fairly simple, or at least compared to the first method. However, if you think it through, you might suspect that the numbers you are multiplying together will not be in natural places in your actual problem.

I want you to try to test this. Write the following on a piece of scrap paper lying around.

       xyz

x     abc

Now, the first step is just cz. Draw a erasable line from z to c. You can see the relationship between the first two numbers being multiplied. Now, erase it.

Here's where it may appear to get awkward. (cy + bz). Draw an erasable line from c to y, and from b to z. What is a easy relationship here?

In fact, you have just made a cross. These two completely random terms have formed a cross on your paper. You can erase the cross now.

Try the third terms, which there are a lot of them! Draw lines between c and x, b and y, and a and z. Believe it or not, you will get a three-way cross, which might resemble an asterisk slightly.

If you continue this process with the other two, you will find just another cross on the left, and then a line on the left side of your problem.

Because of these crosses, this method is called the "Criss-Cross Method." With this crossing pattern in mind, let's try an example. How about 143 x 822.

       143

x     822

If you'll remember, we first have a line on the right side of the problem, connecting the 3 and the 2. So, we must multiply three and two, which will give us six. We then place that directly under the 3 and the 2.

       143

x     822

           6

Now, we must draw our first cross, which is also on the right side of the problem. We have a line connecting the 4 and the 2, and a line connecting the 3 and the 2. Remember that it resembles an X, it is not two parallel lines connecting the four to the left two and the three to the right two. It is the other way around.

Let's find the products of the connected numbers. 4 x 2 = 8 and 3 x 2 = 6. If you'll remember from our algebra, we must add together these products. 8 + 6 = 14.

We cannot stick a 14 there, just like we can't do in any other multiplication method. We will have to put the four, and carry the one.

       143

x     822    1

         46

Now, we have our three-way cross. We must do three computations: 1 x 2, 4 x 2, and 3 x 8. Don't try to do them all at once, just do one at a time. 1 x 2 = 2, and add to that the one we carried to get three. 4 x 2 = 8, and add that to the three to get eleven. 3 x 8 = 24, and add that to the eleven to get thirty-five. We can then put the five and carry the three.

Notice that after each computation, we only had to remember one additional number. This is the power of keeping a running total after each step. The whole purpose of this is to make your final result easier and less messy, which keeping a running total will do.

       143

x     822    3

       546

Now, we will do the next criss-cross, which is the 1 x 2 and the 4 x 8. 1 x 2 = 2, which we can add to the three we carried to get five. 4 x 8 = 32, which we can add to that five to get thirty-seven. We will put the seven, and carry the three.

       143

x     822    3

     7546

Now, we are in home stretch. 1 x 8 = 8, which we will add to the three to get eleven. Since there is nothing to carry over to, we will just put the eleven there.

       143

x     822    3

 117546

Believe it or not, we have just found our answer. We can also continue this up to a 4x4, or a 5x5, or even higher if you want. The only change is the number of criss-crosses you make. For a 4x4, you do your first line, then a cross, then a three-way cross on the right side of the problem, then a four-way cross incorporating all of the numbers in the problem, followed by a three-way cross on the left side of the problem, followed by a two-way cross, followed by that same last line.

This may look a little cumbersome, especially with the gigantic crosses you have to do. However, this method is a lot easier than your traditional multiplication once you start getting into larger factors, and will take much less time to master.

Additionally, it is fun to impress your friends with. If you have them give you a 3x3 multiplication problem, you will be able to knock it out in ten seconds or less, and will not have written a single number on your paper. Since I am known for my mental math, the criss-cross method is something I can use to fool people into thinking I handled the whole problem in my head in one shot, rather than writing down each digit after I compute it.

100th Post: The First Secret of Mental Mathematics

I am thrilled to announce that today marks the 100th post on Cool Math Stuff. It looks like this will also be the longest post, since I have so much I want to say. Anyways, it is really great to be able to talk about what I think makes math cool. But I don't want it to just be what I think, I want you guys to give your input as well. So, please feel free to comment, or send me anything that you find cool about math. I will be more than happy to post about it (it saves me from having to think of something cool to write about).

Before I go into my post, I wanted to start finding out what you think is interesting in math. So, I want you to comment with the most interesting post I have written (the title or link will be fine), and I will take the one with the most comments and either repost it or add some input to it. Here are my top ten favorites, which you can feel free to choose from, or find one of your own.

#1. April 14, 2012; Probability, the Number e, and Magic all in one

http://coolmathstuff123.blogspot.com/2012/04/probability-number-e-and-magic-all-in.html

#2. August 27, 2011; Do the Primes go to Infinity and Beyond?!!

http://coolmathstuff123.blogspot.com/2011/08/do-primes-go-to-infinity-and-beyond.html

#3. July 9, 2011; Divide Almost Any Odd Number into a Number Consisting of all Nines

http://coolmathstuff123.blogspot.com/2011/07/divide-almost-any-odd-number-into.html

#4. March 24, 2012; Pi, Lie, Same Thing... (Part Two)

http://coolmathstuff123.blogspot.com/2012/03/pi-lie-same-thing-part-2.html

#5. September 10, 2011; Another Probability Paradox: What’s Your Birthday?

http://coolmathstuff123.blogspot.com/2011/09/another-probability-paradox-whats-your.html

#6. October 15, 2011; Why Does 64 = 65? Or does it...

http://coolmathstuff123.blogspot.com/2011/10/why-does-64-65-or-does-it.html

#7. June 25, 2011; Greatest Common Factor Made Easy: It's Euclid to the Rescue

http://coolmathstuff123.blogspot.com/2011/06/greatest-common-factor-made-easy-its.html

#8. September 17, 2011; What does .99999... Really Mean?

http://coolmathstuff123.blogspot.com/2011/09/what-does-9999999999999-really-mean.html

#9. July 23, 2011; Patterns and Puzzles at CTY

http://coolmathstuff123.blogspot.com/2011/07/patterns-and-puzzles-at-cty.html

#10. January 14, 2012; Can you correctly add six numbers?

http://coolmathstuff123.blogspot.com/2012/01/can-you-correctly-add-six-numbers.html

Since this is the 100th post, I wanted to make it something that is really special. And in fact, it is what pulled me into the fascinating aspects of mathematics.

In February of 2010, I was shown a video on TED of math professor and mathemagician Dr. Arthur Benjamin, who was doing incredible feats of mental math in his head. As a curious ten-year-old, I of course wanted to be able to do it too. So, my dad bought me his book, Secrets of Mental Math (there is now a fantastic video course on it, which you can click here to see), and I was slowly pulled into the whole mathematics, magic, and science community.

But those first few pages were what really drew me in. So, I wanted my hundredth post to be about what those first few pages were on: how to multiply by eleven in your head.

First of all, lots of you probably remember from third grade when you learned the following pattern:

1 x 11 = 11
2 x 11 = 22
3 x 11 = 33
4 x 11 = 44
5 x 11 = 55
6 x 11 = 66
7 x 11 = 77
8 x 11 = 88
9 x 11 = 99

That's always a fun one. But there still is a pattern that continues. Let's move it up to two-digit numbers.

10 x 11 = 110
11 x 11 = 121
12 x 11 = 132
13 x 11 = 143
14 x 11 = 154
15 x 11 = 165
16 x 11 = 176
17 x 11 = 187
18 x 11 = 198

This one is a little harder to see. Let me make it a little easier to see.


10 x 11 = 110
11 x 11 = 121
12 x 11 = 132
13 x 11 = 143
14 x 11 = 154
15 x 11 = 165
16 x 11 = 176
17 x 11 = 187
18 x 11 = 198


You might notice that going down the list without the bold gives you 123456789. But the important part is the bold. What numbers do you see?

The numbers we were multiplying by! So, the first digit is the first number and the second digit is the second number. That takes care of a lot of it. But what about the middle digit? Do you see any pattern between the outside numbers and the inside number?

1 + 0 = 1
1 + 1 = 2
1 + 2 = 3
1 + 3 = 4
1 + 4 = 5
1 + 5 = 6
1 + 6 = 7
1 + 7 = 8
1 + 8 = 9

The two outside digits actually sum to the middle one. Let's try a couple. 26 x 11.

2 + 6 = 8
286

How about 61 x 11?

6 + 1 = 7
671

Try 72 x 11.

7 + 2 = 9
792

What about 39 x 11.

3 + 9 = 12
3129

Wait, how can that be? 39 x 11 can't be bigger than 72 x 11. We must have messed up.

I neglected to mention before that there is only room for one digit in the middle spot. Since the full 12 doesn't fit there, the two drops in and the one carries over to the 3, making the answer 429.

1
329
429

What is 67 x 11?

1
637
737

Try 96 x 11. This might be confusing, but stick to our rules.

  1
  956
1056

Let's look at three digit numbers. Do you see any patterns with these ones:

132 x 11 = 1452
254 x 11 = 2794
816 x 11 = 8976
427 x 11 = 4697
354 x 11 = 3894

For the first one, you see the one and two on the outsides, just like the two-digit numbers. But where is the three? What are the four and the five for?

Well, 1 + 3 = 4, and 3 + 2 = 5. This goes for all of the other ones.

How about we try a couple:

1  5 3
 \ / \ /
16  83

153 x 11 = 1683

7  2  4
 \ / \ /
79 64

724 x 11 = 7964

8  4  2
 \ /  \ /
8      2
12  6_
92  62

9    8     7
 \   /  \   /
9           7
1(7+1)5
10857

That last one was a little confusing, but you will get the hang of it with some practice. And you can even take this up to four-digits, fives, and more. Let me quickly go over why this works. Let's just do 15 x 11 with classic multiplication from grade school.

    15
x  11

So, 1 x 5 is five.


    15
x  11
      5

1 x 1 = 1. Remember to put the zero down as a place holder.


    15
x  11
    15
      0

Now, 1 x 5 is 5. 1 x 1 is 1.


    15
x  11
    15
  150

Add them up:



    15
x  11
    15
+150
  165

That is great. But, what happened in the process? On the left end, you have just the one (from the one in the fifteen). On the right, you have just a five. In the middle, you have the one and the five, which is the same as adding the two digits together and putting them in the middle.

I find the proof of this pretty cool, but I especially find it cool that you can multiply a number by eleven in seconds with a day or two of practice.

Bonus: I don't normally do a bonus on a long post, but this is something I would really like to share. Recently, I was thrilled to receive an email from a reader of the blog, who wanted to know a formula for figuring out a problem like so (this is his exact words in the email):

"Lets just say I have 30 apples and I want to consume them over a 20 day period.  I want to eat 2 apples a day for as many days as possible; and then cut back to one apple so that I am finishing the last apple on the 20th day."

We can see that you would eat two apples in ten days and one apple for the remaining ten. However, I played around with some numbers and did come up with a formula that will generate the answer to this question.

First, let's set some variables.

a = Total Number of Apples (or object of choice)

d = Total Number of Days (or time interval of choice)

n = Smaller Apple Portion (or object of choice)

x = Number of Days to eat bigger apple portion

To find n, all you have to do is divide d into a and ignore the decimal following. For your example, 30/20 = 1.5, so the smaller apple portion is equal to 1 apple.

To find x, you can use the following formula:

x = a - dn

For your example, you would do:

x = (30) - (20)(1)

x = 30 - 20

x = 10

Ten would be the number of days you eat two apples. To find out how many days you will eat a single apple, just subtract x from the total number of days.

So, if you wanted to eat 50 apples in eight days, you would do the same thing. 50/8 = 6.25, so six is the smaller interval.

x = (50) - (8)(6)

x = 50 - 48

x = 2

You would eat seven apples for two days and six for the rest.

I was really happy to see some feedback and ideas for posts, and I would love for all of you to contribute as well. This is the hundredth post, and I think everything was something that I chose to post. I would really like you guys to email me your favorite cool math thing, and I will be more than happy to post it.

Learn to Add Over Fifteen Numbers in Under Fifteen Seconds

Since I haven't gone over any mental math tricks in an extremely long time, I wanted to get one in this week.

Before this post, I have done posts involving multiplication and squaring tricks. But today, I am going to talk about some addition.

There are ways of doing mental addition reasonably quickly, but some specific series of numbers (like the Fibonacci series, which I taught a way to add last December) can be added much quicker. So, we are going to need some more fastidiousness when we get our numbers to add.

The series we will be adding is the powers of two. Have your volunteer choose a number and list out that many powers of two. Let's say they say nine.

1 + 2 + 4 + 8 + 16 + 32 + 64 + 128 + 256 =

Dont retrogress yet. All you have to do is double the final number in the sequence and then subtract one.

256 x 2 - 1

512 - 1

511

And there is your answer. If you have a little more prowess, you could try going up to sixteen numbers.

1 + 2 + 4 + 8 + 16 + 32 + 64 + 128 + 256 + 512 + 1024 + 2048 + 4096 + 8192 + 16384 + 32768 =

If you do the math, you will find that the answer is 65535.

This method is actually fairly easy to prove. Remember back when we did infinite series? Let's write that one we are subtracting as an infinite series.

2^0 + 2^1 + 2^2 + ... + 2^n-2 + 2^n-1 = 2^n - (2^-1 + 2^-2 + 2^-3 + ...)

Now, add that series to both sides. If we write it as an infinite series, we get:

2^n = 2^n-1 + 2^n-2 + ... + 2^1 + 2^0 + 2^-1 + 2^-2 + ...

The left side is already simplified at 2^n. How do we simplify the right side? I may have mentioned in a earlier post (I'm not sure which) how to simplify this type of series. Let me explain it again.

1. Find what number you multiply each number in the sequence by to get the next number in the sequence. Call that s.
2. Call the first and biggest number in the sequence x.
3. Use the formula x/(1-s) to get the total.

In this sequence, we are multiplying by 1/2. 1/2(2^n-1) can be written as 2^n-2.

The biggest number is the 2^n-1. So, we plug it into the formula to get:

(2^n-1)/(1 - 1//2)
(2^n-1)/(1/2)
2(2^n-1)

Using the law of exponents, we can simplify that to get:

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

And there is the proof. Though the proof is kind of cool, I found it cool that you could add this many numbers up so quickly. Good luck fooling people with this one!