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.

*: 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 one__*: 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 two__*: 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 three__*: 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 four__*: 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 five__*: 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 six__*: 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.*

__Divisibility by seven__
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.

*: 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*

__Divisibility by eight__*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.

*: 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.*

__Divisibility by nine__
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!

*: 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 ten__*: 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.*

__Divisibility by 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.

10

^{0}– (0 • 11) = 1
10

^{1}– (1 • 11) = -1
10

^{2}– (9 • 11) = 1
10

^{3 }– (91 • 11) = -1
10

^{4}– (909 • 11) = 1
10

^{5}– (9091 • 11) = -1
10

^{6}– (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.

*: 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 twelve__*: 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*

__Divisibility by thirteen or greater (the “create a zero, kill a zero method”)__*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.

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