Maths Coursework

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Introduction:

The task we have been set for this piece of coursework is to investigate numbers presented in a number grid. A box is placed around 4 of the number and the question asked was to find the product of the top left number and the bottom right number. This solution was then taken away from the top right number and the bottom left number like this:

(13 x 22)-(12 x23) = 10

 

Aim:

My aim in this coursework is to experiment using 3 different variables in boxes.

These will be: size, shape and size of grid.

I will investigate boxes that are the same size, bigger, smaller, in different proportions and those that are different in shape.

What I mean by this is that I will take boxes of the following proportions:

2x2, 3x3, 4x4, 5x5, 2x3, 2x4, 2x5, 3x4, 3x5, 4x5.

I will take 4 boxes of each of type of the above that are squares. I will take these results by using the random function on my calculator.  The figure given will be any number in the box.

Investigation:

2x2 boxes:

My first boxes that I will be investigating are the 2x2 boxes of which I am taking 4.

Theses are the boxes I have taken:

1:

(36x45)-(35x46) = 10
2:

(58x67)-(57x68) = 10
3:

(89x98)-(88x99) = 10
4:

(43x52)-(42x53) = 10

Lorenzo Brusini

Proving algebraically the 2x2 boxes:

For my prediction of this the 2x2 boxes, I will begin by representing the left top number as n. I will then add or subtract from n to suit the other number in the grid, like this:

As discovered before the answer in the 2x2 boxes always equals 10. If we take these expressions and put it into the same formula as before we should get 10.

(N+10)x(N+1)-N(N+11)= 10

(N2+N+10N+10)-(N2+11N)= 10

N2+11N+10-N2-11n= 10

Therefore 10 = 10

This shows that this is correct for 2x2 boxes.

3x3 boxes

I will now investigate 3x3 boxes that I am once again taking 4 of:

(60x78)-(58x80)= 40

(23x41)-(21x43)= 40

(44x62)-(42x64)= 40

(28x46)-(26x48)= 40

Proving algebraically the 3x3 boxes:

As the answers are the same as before i.e. they all equal the same answer but the answer is higher. This means that it probably can use the same sort of formula.

I shall try to use the formula:

(N+2)x(N+20)-N(N+22)= 40

(N2+20N+2N+40)-(N2+22N)= 40

N2+22N+40-N2-22N= 40

Therefore 40 = 40

Lorenzo Brusini

4x4 boxes

(14x47)-(17x44) = 90

(26x59)-(29x56) = 90

(66x99)-(69x96) = 90

(62x95)-(65x92) = 90

Proving algebraically the 4x4 boxes:

The answers for 4x4 boxes all answer to 90. I shall try to use the algebraic box once again:

Join now!

I shall try to use the formula:

(N+3)x(N+30)-N(N+33) = 90

(N2+30N+3N+90)-(N2+33N) = 90

N2+33N+90-N2-33N = 90

Therefore 90 = 90

5x5 boxes

(55x91)-(51x95) = 160

(26x62)-(22x66) = 160

Lorenzo Brusini

(10x46)-(6x50) = 160

(40x76)-(36x80) = 160

Proving algebraically the 5x5 boxes:

The answers for 5x5 boxes all answer to 160. I shall try to use the algebraic box once again:

I shall try to use the formula:

(N+4)x(N+40)-N(N+44) = 160

(N2+40N+4N+160)-(N2+44N) = 160

N2+44N+160-N2-44N = 160

Therefore 160 = 160

NxN ...

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