• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month
Page
1. 1
1
2. 2
2
3. 3
3
4. 4
4
5. 5
5
6. 6
6
7. 7
7
8. 8
8
9. 9
9
10. 10
10
11. 11
11
12. 12
12
13. 13
13

# Math Portfolio higher level type 1

Extracts from this document...

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  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 expression. Now we want to see if the expression is always divisible by the corresponding.

The first case if =2.

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

Now let's take as a common factor. The expression will become  and since the expression or  it is therefore divisible by 2.

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

Let

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

And 20 is divisible by 2.

Again 182 is divisible by 2.

Also 280 is divisible by 2

Which is divisible by 2

Therefore  is divisible by 2.

Now let's take the second case when

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

Let's take as a common factor the expression will now look like this:

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

Therefore,  is divisible by 3.

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

Let

Now using GDC substitute the following values of in the expression

Which is divisible by 3

And this number is divisible by 3

Middle

is true.

However, we need to prove that   is true.

And .

Now let's subtract from.

The expression is now

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

By collecting terms and simplifying the

Expression it will become:

And by taking as common factor the expression is now:

Therefore it is divisible by 3

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

The 3rd case is when

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

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

Therefore divisible by 5.

Now we assume that is true so is true.

However, we need to prove that   is true.

And .

Now let's subtract from.

The expression is now

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

By collecting terms and simplifying the

Expression it will become:

And by taking as common factor the expression is now:

Therefore it is divisible by 5.

Now let's explore more cases for and if  is divisible bywe'll prove it by induction. So we will factorize the expression  for

Let's look at when

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

Now let's keep factorizing the expression more

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

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

which is not divisible by 6

Which is not divisible by 6.

Conclusion

th row you plug 5 in the expression  instead of so the expression is now by simplifying the expression it will look like this: and 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 , if the number of rows is a prime number the entries in the row will be divisible by . 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 if is a prime number then the expressionis divisible by. Therefore, from this relationship we can conclude that is a multiple of if is a prime number.

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

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

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

which is divisible by 4.

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

This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.

## Found what you're looking for?

• Start learning 29% faster today
• 150,000+ documents available
• Just £6.99 a month

Not the one? Search for your essay title...
• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month

# Related International Baccalaureate Maths essays

1. ## Extended Essay- Math

-ï¿½ï¿½ï¿½J.ï¿½!ï¿½ï¿½<ï¿½1/43/4 ï¿½Sï¿½ï¿½-,ï¿½ï¿½ï¿½ï¿½Fï¿½Tï¿½T0E"ï¿½kï¿½%5ï¿½ï¿½2ï¿½ï¿½Dï¿½ï¿½4ï¿½ï¿½i mUW ï¿½ï¿½M%ï¿½Sï¿½.ï¿½uFßVï¿½!ï¿½ ï¿½ï¿½QO"qï¿½%6hvï¿½6ï¿½"ï¿½@Nï¿½pï¿½,Vï¿½|p"ï¿½ï¿½ ^ï¿½:ï¿½ï¿½7ï¿½(r)ï¿½ï¿½~ï¿½ï¿½D{ï¿½ 3eï¿½'\ï¿½Wï¿½ï¿½,r31EYï¿½#baï¿½Eï¿½2aï¿½-w*ï¿½Kï¿½ï¿½pï¿½ï¿½lï¿½{ï¿½ï¿½ï¿½Ebï¿½*ï¿½ï¿½ï¿½,Øï¿½ ï¿½ï¿½fï¿½3ï¿½ï¿½Æ(c)ï¿½ï¿½)ï¿½X-ï¿½ï¿½]_ï¿½ï¿½9*)3/47ï¿½yï¿½ï¿½ï¿½Yï¿½\$ï¿½ï¿½yJiï¿½hï¿½ï¿½Mï¿½ï¿½,E7oï¿½ï¿½#" Pnï¿½Cï¿½"b' dï¿½ï¿½3/4mï¿½+rï¿½ï¿½Ú.ï¿½nWH#ï¿½<ï¿½(c)ï¿½ï¿½ï¿½0Gï¿½ï¿½ï¿½QTï¿½ï¿½GQï¿½ï¿½m s-Z3/4Xï¿½pï¿½!'<(c)ï¿½"QXÛï¿½ï¿½Dï¿½LjBo w"" D 'ï¿½ï¿½bï¿½ 5ï¿½ï¿½ï¿½ï¿½qQ?T Gï¿½Xï¿½oWï¿½r3/4'ï¿½ï¿½NS q(-+'6ï¿½81/4ï¿½ï¿½ï¿½ï¿½yï¿½FÅª*Rï¿½<ï¿½Xï¿½ÔT-e'nï¿½5>Wï¿½o(r)}'-ï¿½1/23/4Zz/â3/4ï¿½ï¿½Øï¿½`ï¿½j..."ï¿½Pï¿½ ï¿½_^aï¿½ï¿½ï¿½tLa0x0~3jbqUï¿½pï¿½Mï¿½Txi.ï¿½ï¿½5ï¿½ s\$ï¿½ï¿½ï¿½qE1ï¿½5ï¿½ï¿½>vï¿½3ï¿½Mï¿½ï¿½\$A53ï¿½(c)ï¿½(,wï¿½<3/4Fï¿½dï¿½ï¿½qqznï¿½yï¿½ï¿½7#Qï¿½(c)Y-ï¿½)PY=t?0ï¿½ï¿½eE-ï¿½ï¿½]Wï¿½ Xï¿½ptï¿½&ï¿½9(c)1qï¿½,ï¿½ï¿½Í5?ï¿½-ï¿½Åm,ï¿½1#iï¿½ï¿½2Rï¿½h1/4hï¿½L Cï¿½ ï¿½:ï¿½(r)-ï¿½ï¿½ï¿½ï¿½ï¿½"ï¿½Y -&S[ï¿½;|ï¿½ 0G ï¿½Iï¿½ï¿½ï¿½Oï¿½7VKï¿½ï¿½ï¿½ ï¿½qï¿½fï¿½Bï¿½ï¿½*,ï¿½}ï¿½<ï¿½3/4ï¿½@y0--ï¿½sV3/4(tm)_ï¿½2>ï¿½0w![ï¿½_^ï¿½ ï¿½<L)ï¿½`ï¿½Lï¿½hbm'}?ï¿½ï¿½"Êï¿½zfï¿½(r)_ï¿½Jï¿½ï¿½c3/4ï¿½ï¿½Ýï¿½ 'ï¿½j\$Rdsï¿½ï¿½ï¿½(r)ï¿½ï¿½<ï¿½ï¿½ï¿½RPï¿½(tm)U7*ï¿½#ï¿½-ï¿½7[ï¿½qï¿½!ï¿½ï¿½M%ï¿½ï¿½ B,ï¿½Wï¿½-=Gï¿½#"/#ï¿½30ï¿½} Hï¿½}ï¿½6ï¿½ ï¿½3/4ï¿½[ï¿½RV"ï¿½3/4ï¿½+Vï¿½Z)uï¿½-ï¿½ï¿½!ï¿½3ï¿½ï¨ï¿½Cï¿½uï¿½.bï¿½9ï¿½ï¿½ß£MWï¿½jï¿½ï¿½ï¿½ï¿½ï¿½RDUWï¿½b3/42ï¿½ï¿½hï¿½u"ï¿½ï¿½'~ï¿½ï¿½1/2ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½bMï¿½}Umlï¿½ï¿½rYï¿½ï¿½ï¿½-ï¿½ï¿½Aï¿½ï¿½ï¿½ï¿½.ï¿½ï¿½ï¿½Bï¿½eï¿½hï¿½ß5ï¿½1/2(r)ï¿½P+ï¿½ï¿½Xï¿½_AKï¿½`>ï¿½R1*/~)i:ï¿½0ï¿½EK"ï¿½Ë¨ï¿½-ï¿½|iï¿½ï¿½-Oeï¿½eï¿½Y*ï¿½{| 81/2ï¿½Ñ _ï¿½ï¿½iï¿½ï¿½ï¿½~ï¿½; T ï¿½aï¿½ï¿½ï¿½ ï¿½ä´5Îï¿½X}ï¿½ï¿½I"cHTï¿½H>-ï¿½ï¿½ï¿½cAx-Eï¿½"ud((c)ï¿½-ï¿½ï¿½ï¿½N}H"ï¿½ï¿½"Eï¿½iï¿½)(ï¿½fHOï¿½E8tï¿½Rï¿½\$ ï¿½"ï¿½(tm)ï¿½)oH|ï¿½ï¿½"ï¿½Jï¿½_ï¿½ï¿½'ï¿½"(ï¿½ï¿½ Tï¿½ï¿½%(tm)ï¿½ï¿½ï¿½...<Dqaï¿½V`+ï¿½4ï¿½+ ï¿½uï¿½lï¿½ï¿½Oy^L8DPï¿½F0j6dï¿½\ ï¿½N-ï¿½}X%@ï¿½#dï¿½ #!

2. ## Math Portfolio Type II

Such an increasing and decreasing trend in the population is observed in the following years although the population seems to come closer to the sustainable limit, i.e 60000 all the time. In the 8th year, the population is exactly at its sustainable limit and henceforth, the population stabilizes by only varying slightly until the 18th year.

1. ## Math IA type 2. In this task I will be investigating Probabilities and investigating ...

4 points and by at least 2 points, but to save court time, no game is allowed to go beyind 7 points. This means that if deuce is called and each player has 3 points then the next point determines the winner.

2. ## Math Portfolio Type II Gold Medal heights

To provide a reasonably sized and proportioned graph the y-values starting with 10 to 170 were skipped, the (SQUISHY LIGHTENING THINGGY) represents a summarization of these values. The Olympics were not held in 1940 and 1944 as the whole world was caught up in the Second World War.

1. ## SL Math IA: Fishing Rods

10 23 38 55 74 96 120 149 Original ? Distance from Tip (cm) 10 23 38 55 74 96 120 149 New values for the distance from tip were rounded to zero decimal places, to maintain significant figure ? the original values used to find the quadratic formula had zero decimal places, so the new ones shouldn?t either.

2. ## Fishing rods type 2 portfolio

On the other hand, my cubic equation is a polynomial regression. It is a polynomial regression because 1 used all 8 points and not just the 4 required points to determine the model equation. The graph above also indicates that the cubic function is more accurate.

1. ## maths portfolio 1

=6 2x+8=6 2x=-2 x=-1 Hence again I found that again the point of intersection of the linear equations is (-1, 2) The third example would be 4x+11y=18 10x+y=-8 In this case we are suppose to multiply the second equation by 11 to make both the y terms equal 4x+11y=18 110x+11y=-88

2. ## Lacsap triangle investigation.

1st element should have the denominator 11 and the 2nd element should have a denominator of 9. (Excluding the first and last elements of the row valued at ?1?) 1st element =15 - 1(5-1) =15 ? 4 =11 2nd element =15 - 2(5 - 2)

• Over 160,000 pieces
of student written work
• Annotated by
experienced teachers
• Ideas and feedback to