Cool Divisibility Stuff (There's a lot!!)

This week, we are going to talk about divisibility, a pretty broad area of mathematics that we were lucky to be introduced to in fifth grade. I’ve looked into it more, and even use it for a lot of tricks in math.

There are a bunch of tricks for it, and I will explain them all and prove them to you. This post will almost be a lesson, but all of the rules are very cool! Let’s go through them all.

Divisibility by one: If the number has no decimal, it is divisible by one. For these rules, we are only working with decimals. To prove it, we look at the Closure Property, which states that any number multiplied by one will remain the same number. This means every integer is a multiple of one.

Divisibility by two: If the number ends in two, four, six, eight, or zero, it is divisible by two. The definition of even number is, “A natural number that is divisible by 2.” It is also defined as, “A whole number that has 0, 2, 4, 6, or 8 in the ones place.” Since these two things are the same, this means that this rule is valid.

Divisibility by three: This rule is probably one of the coolest, and most magical. If you add up the digits in the number and this gives you a multiple of three, this means that the number is a multiple of three. If the sum is too big for you to know if it has a three factor, go ahead and add up the digits again until you do know. I will prove this one along with the divisibility by nine rule, as it is very similar, and a number that is divisible by nine is also divisible by three.

Divisibility by four: Look at the number’s last two digits, and completely ignore the rest. If the number you are looking at is a multiple of four, then the whole number is a multiple of four. If you think about it, every number is 100x + y with y being the last two digits. Since we know 100x is a multiple of 4 (25x • 4 = 100x), all that leaves us with is y. This means that if y is a multiple of 4 also, then adding 100x won’t change this.

Divisibility by five: If the number ends in five or zero, it is a multiple of five. If you think about it, the digits from 0 to 9 end in either five or zero when multiplied by five. Since the last digit of a multiplication problem is the last digit of the last two digits multiplied together, we are left with all of the integral numbers ending in five or zero.

Divisibility by six: Since six is the product of three and two, we can just put both of these rules to use since they share nothing in common. If the number follows through with the divisibility by three rule (just adding to a multiple of three, no need for a multiple of six, 1 + 2 = 3 and 12 is a multiple of six), and is even, it is a multiple of six. Check the divisibility by two and three rules for more.

Divisibility by seven: This is also a really cool one, one my teacher didn’t even know. Take the last digit of the number and double it. Then, subtract that from the rest of the number. Keep repeating this until you have a one or two digit number (negatives are okay, you can just turn them positive). If the number you have is a multiple of seven, the original number is a multiple of seven. Let’s do one real quick, the number 224.

First, double the 4 to get 8. Subtract 8 from 22 to get 14. Since 14 is definitely a multiple of seven, the 224 is as well. 224 happens to be 32 • 7, so we did our math correctly. To prove this, we will use a little bit of Algebra. Our formula states that if 10r + x is a multiple of seven, then r – 2x is a multiple of seven. Let’s say the number is a multiple of seven. This means that r – 2x is a multiple of seven.

r – 2x = multiple of seven

Let’s try multiplying both sides by ten.

r – 2x = multiple of seven

10(r – 2x) = multiple of seven (and seventy)

10r – 20x = multiple of seven (and seventy)

Forget it is a multiple of seventy. We are not looking for that. However, let’s go in a different direction and add 21x to both sides. Since 21 is a multiple of seven, we still have a multiple of seven on the right-hand side of the equation. However, the left becomes:

10r – 20x = multiple of seven

10r – 20x + 21x = multiple of seven

10r + x = multiple of seven

And this brings us back to where we started. It is a little complicated, but I think it is pretty cool.

Divisibility by eight: Similar to divisibility by four, we will be looking at a block of digits on the right. However, note the fact that eight isn’t a multiple of one hundred. However, it’s a multiple of one thousand. So, we will look at the last three digits and see if it is a multiple of eight. Since this might be a little hard for you, as I don’t have my eights memorized that far, you can look at the hundreds place and see if it is odd or even. If it is even, this means that you can check the last two digits for divisibility by eight because 200 is a multiple of eight. If it is odd, check the last two digits for divisibility by four, but then make sure it is not a multiple of eight, or a multiple of eight plus four. Since 104 is a multiple of eight, this should work as well.

You can even take these principles to divisibility by any power of two. The power you are raising two to is the amount of digits you need to test at the end. So, for divisibility by 64, just check to see if the last six digits are divisible by 64 because 2^6 = 64.

Divisibility by nine: This is unquestionably the most magical of all of the proofs, and is used for so many things. It is the answer to numerous magic tricks, math tricks, and even provides loads of ways to check answers, including divisibility, digital roots, and mod sums. In fact, you can even cube root numbers based on this rule. To find divisibility by nine, you simply add up the digits of the number, and if that is a multiple of nine, the whole number is a multiple of nine. To prove it, let’s test the number 234. 2 + 3 + 4 = 9, so it is definitely a multiple. But why? Let’s write the number 234 in expanded notation, something you learn around third or fourth grade.

2 x 100 + 3 x 10 + 4 x 1

We can rewrite these to get:

2 x (99 + 1) + 3 x (9 + 1) + 4 x (0 + 1)

Let’s distribute the 2, 3, and 4 into the groups of numbers beside them.

(2 x 99) + 2 + (3 x 9) + 3 + (4 x 0) + 4

We can rearrange this to get:

[(2 x 99) + (3 x 9) + (4 x 0)] + 2 + 3 + 4

You might notice, but the sum inside of the brackets is definitely a multiple of nine, as it is being created by all multiples of nine. So, we can eliminate it to get the sum of all of the digits. Here, I don’t think the proof lives up to the magic in the actual trick, but it is pretty cool!

Divisibility by ten: To test divisibility by ten, all we need to do is see what the last digit is. If it is a zero, then the number is a multiple of ten. To prove it, we look at the Power of Zeros rule, which says that to multiply a number by a power of ten, just tack on that power amount of zeros. This means that a number multiplied by 10 has one zero at the end. You can elevate this to say that a multiple of 100 has two zeros at the end, a multiple of 1000 has three and so on.

Divisibility by eleven: To test divisibility by eleven, alternately subtract and add the digits and if you end with a multiple of eleven (it may be zero or negative), then your original number is a multiple of eleven. So for the number 1353, you would go 1 – 3 + 5 – 3 = -2 + 5 – 3 = 3 – 3 = 0. Since zero is a multiple of eleven, the full number is a multiple of eleven.

To prove it, we know that if you subtract eleven from a number constantly, the number still keeps its status as a multiple of eleven. Therefore, let’s put ten to its powers (creating the place values) and see what happens when you constantly subtract multiples of eleven.

10⁰ – (0 • 11) = 1

10¹ – (1 • 11) = -1

10² – (9 • 11) = 1

10³ – (91 • 11) = -1

10⁴ – (909 • 11) = 1

10⁵ – (9091 • 11) = -1

10⁶ – (90909 • 11) = 1

and so on…

This convinces me enough, but I’m not sure of where to take it from there. If you want to add anything, please comment. It would be interesting for all of us.

Divisibility by twelve: Let’s take two distinct ones, divisibility by four and three. If the number is divisible by four and its digits add to a multiple of three, it is a multiple of twelve. Like the divisibility by six rule, this works as well. You can even take this principle to other numbers like this, like 14, 15, 18, 21, 22, 24, 28, and so on.

Divisibility by thirteen or greater (the “create a zero, kill a zero method”): Since this is a little more complicated, let’s learn it through an example. Is 2756 divisible by thirteen? First, we must look at the last digit of 2756, six. We need to find a multiple of thirteen that ends in six. If there is one, that means the number is not divisible, and you have learned a new property about that number. However, 13 x 2 = 26, so we have found one. We must subtract the 26 from 2756 to create a zero. So, 2756 - 26 = 2730. Now, we kill the zero, or just ignore it. Now, we create a zero by subtracting 13 (273 – 13 = 260). After killing it, we have 26. Since 26 is a multiple of 13, 273 and more importantly, 2756 are multiples of 13. You could even have subtracted 156 from the original number if you knew that it was 12 x 13. This would make it only take two steps instead of three.

Even though it is a lot in one dose, it is definitely worth practicing. These are really cool principles, and you could probably figure out a few tricks out of them. If you watch my Mathemagics show closely, I will use divisibility to perform one of the tricks, so you can investigate that as well.

How many cube roots of one are there?

When getting into algebra, you will hear the term "real number" a lot, without an explanation as to why you can't just put "number." Since no one asks the question, it just slides by, and when the question is asked, the teacher responds, "You'll learn that in Algebra II," or something along those lines.

The answer to that question is that there is such thing as "imaginary numbers," or complex numbers, which are made up of a constant, a coefficient, and the letter i, which symbolizes the square root of negative one.

What is the square root of negative one? Some say negative one. Well, (-1) x (-1) = 1, so that is incorrect. People then turn around and say one. Well, 1 x 1 = 1, so that is incorrect. Then, they might try 1/2. Well, 1/2 x 1/2 = 1/4, so that is wrong. They will keep trying things until they give up.

What is the answer? If you think about it, a negative times a negative, or a negative squared, is a positive. A positive times a positive, or a positive squared, is a positive. So, you cannot square a real number and get a negative. So, mathematician Heron of Alexandria came up with the letter i, and began using that as the square root of -1. So, then, by the Multiplication Property of Square Roots, you can conclude that √(-9) would be 3i because you can break that into √(9)√(-1) = 3√(-1) = 3i. Rafael Bombelli built on his works, and make this concept a regular part of Algebra.

Now, let me introduce you to one more thing about imaginary numbers before I show you about the cube rooting. When you write these terms out, you write like 5 + 3i, which means 5 + √(-9), just like how you'd write a real square root in Algebra. The 5 + 3i would be known as a complex number, which is a number involving i. The conjugate of a complex number is to keep the same expression, but switch the operation separating them. For instance, the conjugate of 5 + 3i = 5 - 3i because we kept the same term, but switched the operation.

If you think about it, a complex number's conjugate is equal to the number because any number has two square roots, a positive one and a negative one. So by making the i term negative, we are just looking at the other root.

Now that we've gotten that out of the way, let's get to the good part! Let's take the complex number -1/2 + i/2√(3). It is the same thing, with a constant of -1/2 and coefficient of 1/2√(3). How about we cube it.

(-1/2 + i/2√(3))(-1/2 + i/2√(3))(-1/2 + i/2√(3))

First, we'll square it. We can use FOIL for that. If you don't know, it stands for "First, outer, inner last." It's basically the distributive property made simpler for multiplying binomials.

(-1/2 + i/2√(3))(-1/2 + i/2√(3))
1/4 - i/4√(3) - i/4√(3) - 3/4
-1/2 - i/2√(3)

We ended up with the conjugate of before. That's interesting. Let's finish off by multiplying by the -1/2 + i/2√(3).

(-1/2 - i/2√(3))(-1/2 + i/2√(3))
1/4 - i/4√(3) + i/4√(3) + 3/4
1 - i/4√(3) + i/4√(3)

What do we do with that? Well, there are i's in both terms, so we can combine them. However, look closer. They are opposites of each other, or the additive inverse of each other. What does that mean? The definition of additive inverses are two numbers in which when added together give you zero. So, these two confusing numbers simplify to zero!

1 - i/4√(3) + i/4√(3)
1 + 0
1

So, we are left with 1 as our answer! We did nothing wrong there. -1/2 + i/2√(3) is in fact the cube root of one, as well as its conjugate and of course, the integer one. Was this random? No! Mathematics is never random!

If you take the Cartesian Plane, and make the numbers going up the y-axis i, 2i, 3i, 4i, 5i, etc. and -i, -2i, -3i going down, you have the Imaginary Cartesian Plane. If you make a circle going through the points (1,0), (0, i), (-1, 0), and (0, -i), then you will have a unit circle. To find the 1st root of one, we of course start at (1, 0) and that is it. For the square root, or the second root, we would split the 360° of the circle in half to get 180°. So, we have the 1, and then we travel 180° to get -1, the other square root. For the fourth root, we could split 360 in fourths to get 90°, and at ninety degrees, all of the roots are found, 1, -1, i, and -i.

What about if we split in thirds, or 120°. Then, we end up at the points 1, -1/2 + i/2√(3), and -1/2 - i/2√(3). You can check that if you'd like. At the 72 degree marks, you will find the fifth roots, and the 60 degree marks give you the sixth roots.

If you know anything else about this, please tell us! Also, we will probably be taking more about imaginary numbers, so if you want me to show anything in particular, let me know.

Why Does 64 = 65? Or Does It...

Why does 64 = 65? What kind of a question is that? Any pre-schooler probably knows that 64 doesn't equal 65. Algebra clearly shows that they are not equal. However, geometry might throw us off track.

Let's take a chessboard. It's an 8 x 8 grid, with a total area of 64 square units. I'd like you to make the following cuts in the chessboard, as well as along the 5th row from the top (3rd row from the bottom).

Now, arrange these shapes into a rectangle. Let's check out its dimensions. We have a side that is 5 units long, and a side that is 13 units long. To figure out the area of the rectangle, we would do 5 x 13 = 65.

Not good enough? Make the shapes into a triangle. We have 10 units for the base, and 13 units for the height. To find area, we do (bh)/2. So, 10 x 13 = 130 ÷ 2 = 65.

How is this possible? We have taken a grid with area 64 and just by rearranging the shapes, end up with a grid with area 65. No, there was no human error involved, your cuts probably were very accurate, and even a perfectly straight cut would still give you 65 as your area. So, how is this possible?

Even I completely understand that this is completely invalid. However, I still can't wrap my head around why it is wrong. I've been told that the squares along the cut after you assemble the shape are not valid squares, which is the only thing that seems accurate. However, with this, I just find it fascinating to take knowledge you learned in kindergarten, or even pre-school, is being challenged with this very contradictory proof.

I also saw this as a proof of last week's maneuver with Fibonacci numbers. I am not quite sure of why, but it does make some sense, that you are turning eight squared into thirteen times five. It does work again with a 5x5 or 13x13 grid, as long as you make the correct cuts. 

Region Revenge: In August, I gave you guys a problem called region revenge. The goal was to find its explicit formula. The answer 2^n-1 is incorrect, as the sixth cut can only make 31 regions, the seventh makes 57, and so on. Here is the correct answer:

An = (n^4 - 6n^3 + 23n^2 - 18n + 24)/24

You could have solved this with the techniques we used for the other problems, just by creating a five way system.

Fibonacci Day: More Patterns in the Squares

Today yet again is a Fibonacci Day! It is October 8th, and 8 is the sixth Fibonacci number. To keep the theme of last week, let's use the square Fibonacci numbers again. Here they are:

1     1     2     3     5     8     13     21     34
1     1     4     9   25   64   169   441 1156

Let's take each Fibonacci number and move one away from it. Now, we'll multiply those numbers and see how close we get to the square.

1) 0 x 1 = 0 = 1^2 - 1
2) 1 x 2 = 2 = 1^2 + 1
3) 1 x 3 = 3 = 2^2 - 1
4) 2 x 5 = 10 = 3^2 + 1
5) 3 x 8 = 24 = 5^2 - 1
6) 5 x 13 = 65 = 8^2 + 1
7) 8 x 21 = 168 = 13^2 - 1

And so on and so forth. Basically, Fn-1 x Fn+1 = Fn^2 ± 1, or even more accurate, Fn-1 x Fn+1 = Fn^2 + (-1)^n. Let's try it again, this time looking two away from the number. Keep in mind that 1 is the negative-first Fibonacci number.

1) 1 x 2 = 2 = 1^2 + 1
2) 0 x 5 = 0 = 1^2 - 1
3) 1 x 5 = 5 = 2^2 + 1
4) 1 x 8 = 8 = 3^2 - 1
5) 2 x 13 = 26 = 5^2 + 1
6) 3 x 21 = 63 = 8^2 - 1
7) 5 x 34 = 170 = 13^2 + 1

This time, we have pretty much the same pattern. Fn-2 x Fn+2 = Fn^2 ± 1, or  Fn-2 x Fn+2 = Fn^2 - (-1)^n. How about we move three away. It's the same type of pattern, but a little different. Keep in mind that -1 is the negative-second Fibonacci number (since a Fibonacci number is the two numbers before it added together, than the zero comes from x + 1, or -1 + 1).

1) -1 x 3 = -3 = 1^2 - 4
2) 1 x 5 = 5 = 1^2 + 4
3) 0 x 8 = 0 = 2^2 - 4
4) 1 x 13 = 13 = 3^2 + 4
5) 1 x 21 = 21 = 5^2 - 4
6) 2 x 34 = 68 = 8^2 + 4
7) 3 x 55 = 165 = 13^2 - 4

We have the same idea. We are stuck with a four, giving us the pattern of  Fn-3 x Fn+3 = Fn^2 + 4(-1)^n. Let's look at our neighbors four away and see if we can see the pattern better. What do you think the negative-third Fibonacci number is? If you got two, then good job.

1) 2 x 5 = 10 = 1^2 + 9
2) -1 x 8 = -8 = 1^2 - 9
3) 1 x 13 = 13 = 2^2 + 9
4) 0 x 21 = 0 = 3^2 - 9
5) 1 x 34 = 34 = 5^2 + 9
6) 1 x 55 = 55 = 8^2 - 9
7) 2 x 89 = 178 = 13^2 + 9

Same idea again. We have Fn-4 x Fn+4 = Fn^2 - 9(-1)^n. However, the four and nine aren't there randomly. Let's look these differences closer.

1, 1, 4, 9

Recognize them? They are the squares of the Fibonacci numbers again! If you go five away, it is the square of the fifth Fibonacci number, six away is the square of the sixth Fibonacci number, one hundred away is the square of the hundredth Fibonacci number. Basically, a general formula is Fn-a x Fn+a = Fn^2 ± (Fa^2)(-1)^n. Or, you can use the below formula to be even more accurate:

Fn-a x Fn+a = Fn^2 - ((-1)^a)(Fa^2)((-1)^n)

I have no clue why this works, but please put up a proof if you know it. This is one of the coolest things about Fibonacci numbers!

Fibonacci Day: Adding the Squares

Today is a Fibonacci Day! It is the first, and one is a Fibonacci number. One is in fact two Fibonacci numbers. In the last post, you learned how to square numbers that end in five. To keep this squaring theme, how about we square the Fibonacci numbers.

1, 1, 2, 3, 5, 8, 13, 21, 34...
1, 1, 4, 9, 25, 64, 169, 441, 1156...

We've been doing lots of adding Fibonacci numbers. Let's finish it off by adding the square Fibonacci numbers.

1 = 1
1 + 1 = 2
1 + 1 + 4 = 6
1 + 1 + 4 + 9 = 15
1 + 1 + 4 + 9 + 25 = 40
1 + 1 + 4 + 9 + 25 + 64 = 104

Do you see a pattern? It is a little hard to find, but definitely present. Look at this:

1 = 1 =                                       1 x 1

1 + 1 = 2 =                                 1 x 2

1 + 1 + 4 = 6 =                           2 x 3

1 + 1 + 4 + 9 = 15 =                   3 x 5

1 + 1 + 4 + 9 + 25 = 40 =           5 x 8

1 + 1 + 4 + 9 + 25 + 64 = 104 = 8 x 13

They are the product of the two consecutive Fibonacci numbers! Why on earth would that be? I had recently looked for one on the internet, and found an amazing geometric proof for it.

Have you ever heard of the golden rectangle, or the golden ratio? We touched on the golden ratio when I gave you the explicit formula for Fibonacci numbers (the golden ratio is the same as the greek letter fi). The golden rectangle is a rectangle of which the ratio of the length and width is the golden ratio. Something else whose ratio is the golden ratio is Fibonacci numbers! So, the side lengths of the rectangle are consecutive Fibonacci numbers!

Since we are dealing with squares of Fibonacci numbers, let's make some squares.

We've just taken these squares and organized them in a fashion that makes the side lengths two Fibonacci numbers. We went up to 34 squared, so let's see what the side lengths are.

They are 34 and 55. So, to figure out the area of the whole thing, you can add up the areas of all the squares, or just multiply the 34 by 55. And because of the way it is laid out, you can do it with any Fibonacci numbers! I think that is really cool!

Bonus Pattern: How about we add the squares of consecutive Fibonacci numbers.

1 + 1 = 2

1 + 4 = 5

4 + 9 = 13

9 + 25 = 34

25 + 64 = 89

64 + 169 = 233

The sums of the consecutive square Fibonacci numbers is in fact a Fibonacci number. I don't know a proof for this, but please tell me if you find one!