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Math Portfolio higher level type 1

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Introduction

Investigating Divisibility

In order to determine if an expression is divisible by a certain value, we factorize the expression and see if we can take the corresponding value let's call it image00.png as a common factor. Afterwards, we see if it is divisible depending on how the expression will turn out. I'll explain more with examples.

Now let's look at the expressionimage01.pngimage61.pngimage71.png. Now we want to see if the expression is always divisible by the correspondingimage00.png.

The first case if image00.png=2.

Now by substituting 2 in the expression, the expression will look like this:

image87.pngimage61.png

Now let's take image09.pngas a common factor. The expression will become image103.png and since the expressionimage02.png or image10.png it is therefore divisible by 2.

Now let's check the validity of my statement let's take a few examples.

Let  image18.png

Using GDC we plug the following values in the expression and check if it is divisible by 2.

image26.pngAnd 20 is divisible by 2.

image36.png Again 182 is divisible by 2.

image41.pngAlso 280 is divisible by 2

image49.png Which is divisible by 2

Therefore image03.png is divisible by 2.

Now let's take the second case when image57.png

Now by substituting 3 in the expression it will turn out to be like this:

image60.png

Let's take image62.pngas a common factor the expression will now look like this:

image64.png And now by factorizing it more where image65.pngis difference between two squares, the expression will look like this image66.png which is three successive (consecutive) terms.

 Therefore, image03.png is divisible by 3.

To make sure this is true let's take a few examples.

Let image67.png

Now using GDC substitute the following values of image09.pngin the expression

image68.pngWhich is divisible by 3

image69.pngAnd this number is divisible by 3

image70.png

...read more.

Middle

image23.pngis true.

However, we need to prove that image20.png  is true.

image95.png And image96.png.

Now let's subtractimage23.png fromimage27.png.

The expression is now image97.png

Let's solve the brackets so the expression becomes in this form:

image98.png By collecting terms and simplifying the

Expression it will become:

image99.png

And by taking image100.pngas common factor the expression is now:

image101.png

Therefore it is divisible by 3

Since image03.png is not divisible by 4 it is ignored and we don't have to prove by induction.

The 3rd case is when image79.png

Now by using mathematical induction we want to prove that image20.png is divisible by 5.

First we let n=1 and see if it is divisible by 5

image102.png Therefore divisible by 5.

Now we assume that image22.pngis true so image23.pngis true.

However, we need to prove that image20.png  is true.

image104.png And image105.png.

Now let's subtractimage23.png fromimage27.png.

The expression is now image58.png

Let's solve the brackets, so the expression becomes in this form:

image106.png By collecting terms and simplifying the

Expression it will become:

image59.png

And by taking image107.pngas common factor the expression is now:

image108.png

Therefore it is divisible by 5.

Now let's explore more cases for image00.pngand if image03.png is divisible byimage00.pngwe'll prove it by induction. So we will factorize the expression image01.png for image04.png

Let's look at when image05.png

Now let's plug the value of image00.png in the expression.

image06.png Now let's keep factorizing the expression more

image07.png

I don't find any clear evidence in the expression to show if the expression is divisible by 6 or not however, let's take a few example to check.

Let image08.png

Now by plugging the following image09.pngvalues in the expression we check if it is divisible by 6.

image11.png which is not divisible by 6

image12.pngWhich is not divisible by 6.

...read more.

Conclusion

th row you plug 5 in the expression image20.png instead of image00.pngso the expression is now image58.pngby simplifying the expression it will look like this: image59.pngand 5 10 10 5 are the entries in the fifth row and since each row starts and ends with 1 you can find the row and its entries by this relation. So the fifth row of Pascal's triangle will look like this 1 5 10 10 5 1. In addition, there's another relationship between Pascal's triangle and image20.png, if image00.pngthe number of rows is a prime number the entries in the row will be divisible by image00.png. For example, let's look at the 3rd row of Pascal's triangle. The entries in the 3rd row     1 3 3 1 are divisible by 3 since 3 is a prime number. The same is with the expression image20.pngif image00.pngis a prime number then the expressionimage20.pngis divisible byimage00.png. Therefore, from this relationship we can conclude that image56.pngis a multiple of image00.pngif image00.pngis a prime number.

After finding this relationship I came up with a new conjecture. If image00.pngin the expression image01.png is a prime number then image03.png is divisible by image00.png. Also, if image00.png is a prime number in Pascal's triangle where image00.pngis the number of rows then the entries in the row are divisible by image00.pngand it is a multiple ofimage56.png.

However, if we look at the converse of this statement which is if expression image01.png divides by image00.pngthen image00.png is a prime number. This statement is not true since this expression can divide by some non-prime numbers for some values ofimage09.png.

For example let's take 4, although 4 is not prime yet if you plug 5 in the expression image01.png instead of image09.pngit will divide by 4.

image63.png which is divisible by 4.

Therefore, the converse of my conjecture is not true and doesn't hold.

...read more.

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