Maths coursework number grid
In this project, I am going to investigate a number grid. Using a set of instructions, that, have been given to me. The instructions are to find the product of various numbers. Before I start I will like to explain exactly why I have colour coded my work. If you look at my project you will find that there are certain numbers in colour the reason for this is, that it makes it easier to understand what is being multiplied and what is being subtracted.
The numbers that were initially given to me were 12, 13, 22, 23 presented in a number grid marked by a two by two box.
I was told to find the product of top left, (12) and the bottom right number (23). We then had to do the same to the top right (13) and bottom left (22). Once I we had worked out both products I had to calculate the difference.
I am now going to give two examples to show you what I had to do.
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Example One
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22
23
12 13 DIFFERENCE
x 23 x 22 286
= 276 = 286 - 276
= 10
I would like to show one more example of this to see if the difference will always be 10.
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24
25
14 15 Difference
x 25 x 24 360
350 360 - 350
10
I am now going to use algebra to try and prove that the difference will always be
Ten and that the examples shown above are not just coincidences.
X
X+1
X+10
X+11
Remember that this box is exactly the same as the first but the numbers have been put in to algebraic expressions.
Using my original set of instructions, I am going to work out the algebraic two by two box.
X(11+X) (X+1)(X+10)
=X2+11X =X2+11+1
DIFFERENCE
I am now going to subtract the answers I got from the algebra and see if the difference is 10.
X2+11+1
-X2+11X
=10
I have just used algebra and proved the difference will always be ten.
The algebra was labelled using the terms X, X+1, X+10 And X+11. the point of this was, when we look at the original box that was used in example one, the first number is twelve to replace this, the first number in the algebraic two by two box is going to be X. the second number is 11, which is one unit bigger than twelve therefore the second term which was used for algebra is X+1. In example one the number below the twelve (X) is 22 which is 10 units bigger than twelve therefore the term we used for this, is X+10. Finally the number below 13 (X+1) is 23 which is 11 units bigger than 12 (X) Therefore the term I used for this is X+11.
Using my original set of instructions, I am going to work out the algebraic three by three box.
Note that the numbers not being used are not coloured nor are they under lined
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68
69
Difference
47 49 3283
x 69 x 67 - 3243
3243 3283 40
Like I did before I am now going to use algebra to see if the difference is always
Forty, in a three by three ...
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Using my original set of instructions, I am going to work out the algebraic three by three box.
Note that the numbers not being used are not coloured nor are they under lined
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49
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68
69
Difference
47 49 3283
x 69 x 67 - 3243
3243 3283 40
Like I did before I am now going to use algebra to see if the difference is always
Forty, in a three by three box.
Just as is did before I have only highlighted the expressions being used.
X
X+1
X+2
X+10
X+11
X+12
X+20
X+21
X+22
X(X+22) (X+2)(X+20)
=X2+22X =X2+20X+2X+40
= X2+22X+40
Difference
-X2+22X+40
X2+22X
= 40
The algebra proved me to be right that in a three by three box the difference is always going to be forty.
Now I am going to do one more example before I move onto the next stage of my project, for this example I am going to do a four by four box.
Using exactly the same rules as I have used in my previous examples I am going to try and work out the difference followed by algebra.
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Difference
61 64 5824
x94 x91 -5734
5734 5824 90
X
X+1
X+2
X+3
X+10
X+11
X+12
X+13
X+20
X+21
X+22
X+23
X+30
X+31
X+32
X+33
Difference
X(X+33) (X+3)( X+30) X2+30X+3X+90
=X2+33X =X2+30X+3X+90 X2+33X
=90
Now I am convinced that algebra will give me the difference for various different sized boxes I am not going to try a number term example but skip straight to algebra for a five by five box.
X
X+1
X+2
X+3
X+4
X+10
X+11
X+12
X+13
X+14
X+20
X+21
X+22
X+23
X+24
X+30
X+31
X+32
X+33
X+34
X+40
X+41
X+42
X+43
X+44
X(X+44) (X+4)( X+40) Difference
=X2+44X =X+40X+4X+160 X+44X+160
=X+44X++160 - X2+44X =160
I will use a table to try and work out a formula that will calculate the difference of the squares (boxes) with out using algebra.
Table 1 Table two
box size difference box size difference
2x2 10 2x2 1
3x3 40 3x3 4
4x4 90 4x4 9
5x5 160 5x5 16
If we take all the zeros of the differences in table 1we will get a table resembling table 2. I have now worked out a formula. Looking at table 2 if I want to find out the difference for the 4x4 box all I need to do is subtract one from four (3) multiply that, by one from four (3) and times by ten my formula should look like this. Using 'd', for difference I can present my formula.
d=10(b-1)(b-1)
d=10(b-1)2
d=10(4-1)(4-1)
d=32x10
d=9x10
d=90
This formula proves me right as the difference for a 4x4 box is in fact 40. this formula works and I therefore do not need to use algebra to work out the difference on any sized sqaure box.
I wondered what would happen to boxes representing consecutive multiples.
First I took a two by two square, with consecutive multiple numbers in the two times table.
I am going to compare it with my original two by two square, where the difference is ten, and my formula below confirms that.
d=(b-1)2x10
d=(2-1) 2x10
d=12 x10
d=10
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16 14 Difference
x 4 x6 84
84 64 -64
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X+2
X+10
X+12
X(X+12) (X+10)(X+2) DIFFERENCE
=X2+12X =X2+10X+2X+20 X2+12X+20
= X2+12X+20 -X2+12X
= 20
The original general formula is
d=(b-1)2x10
Using this formula I can now work out a new formula for consecutive multiples. The formula will have one change to it and that will be, instead of multiplying by ten, we will multiply by ten times the multiple. My new formula which I am going to use to calculate the difference of a four by four square, which numbers are in the four times table, using 'm' for multiples my formula should look like this.
four by four two by two
d=10m(b-1)2 d=10m(b-1)2
d=(4-1)2x40 d=(2-1)2x20
d=32x40 d=12x20
d=9x40 d=1x20
d=360 d=20
I am now going to use algebra to see if my new formula was correct to calculate the differences for consecutive multiples.
X
X+4
X+8
X+12
X+10
X+14
X+18
X+22
X+20
X+24
X+28
X+32
X+30
X+34
X+38
X+42
X(X+42) (X+12)(X+30) DIFFERENCE
=X2+42X =X2+30x+12X+360 X2+42X+360
= X2+42X+360 - X2+42X
= 360
Algebra has proved to me, that the formula d=10m(b-1)2 is accurate and I now do not to use algebra to work out the difference on consecutive numbers as I could just use my formula.
I realised I had used in my examples for my new formula of d=10m(b-1)2, that I had used a two by two box multiples of two, and a four by four box multiples of four. So I might be wrong as this formula might only work with boxes where the multiples have to be same as the box size. So I tried a six by six box, multiples of ten.
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70 120 Difference
x70 x20 4,900
4,900 2,400 -2,400
=2,500
Instead if using algebra to check that my calculations are right I am going to use my formula of d=10m(b-1)2.
d=10m(b-1)2
d=10mx52
d=10mx25
d=100x25
d=2500
The difference for my six by six box, multiples of ten, that I got from my formula above, is two thousand five hundred. I got the same answer form my working out above the formula. This shows me that I was wrong in doubting my-self that the formula only works with boxes where the multiples are the same as the box size.
Finally I looked at the rectangle.
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I realised the only difference between the square and rectangle are, that the length and breadth are not equal. However the rules that I was given in my original number grid stay the same.
I called the length 'l' and the breadth 'w'. I am using 'w' for breadth as in my last formula, d=10m(b-1)2 I used 'b', and if I use 'b' for my new formula that I am going to try and work out later in the project, it will start getting confusing.
Using the a rectangle taken from my new number grid (above).
19 16 Difference
x6 x9 144
44 114 -114
30
X
X+1
X+2
X+3
X+10
X+11
X+12
X+13
X(X+13) (X+3)(X+10) Difference
X2+13X X2+10X+30+3X X2+13X+30
X2+13X+30 X2+13X
d=30
I am convinced that there is a formula to work out the difference of a Rectangle.
To make a table I am going to use an eight by five rectangle and a four by seven rectangle.
Eight by five
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75 71 Difference
x1 x5 355
75 355 -75
280
Seven by four
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100 97 Difference
x37 x40 3880
3700 3880 -3700
=180
I made a table realising that differences from my rectangles were multiples of ten. In the first I got thirty. The second example I got two hundred and eighty, and in the third example I got one hundred and eighty.
I made a table using 'l' for length and 'w' for breadth.
RECTANGLE
RESULT
Divide by ten
4x2
30
3
8x5
280
28
7x4
80
8
To obtain a general formula I am now going to make another table similar to the one above.
RECTANGLE
RESULT
Divide by ten
Multiplied by ten
Obtaining the general formula
4x2
30
3
30
d=(l-1)(b-1)x10
8x5
280
28
280
d=(l-1)(b-1)x10
7x4
80
8
80
d=(l-1)(b-1)x10
I think I might now have a general formula. To check whether it is accurate in finding out the difference of a rectangle. I am going to try it using the answers of my last two previous examples. Note that I did not use algebra in those examples so I don't know if that the calculations I did are correct. However if my formula is accurate it will show me if my calculations were correct.
To start of with I will use the eight by five rectangle.
d=(l-1)(b-1)x10
d=10(7x4)
d=10x28
d=280
Just to be sure that this calculation is not a coincidence I am going to use the same formula for my seven by four rectangle. If the difference turns out to be the same as my original calculation of the seven by four rectangle, I can be sure that the formula works, and can be used as a general formula.
d=(l-1)(b-1)x10
d=10(6X3)
d=10x18
d=180
I have now proved that my formula, d=(l-1)(b-1)x10, has worked and will calculate the difference of any sized rectangle. I now do not need to use algebra or number term calculations to find the difference of a rectangle.
For the last part of my project I looked at consecutive multiples of the rectangle.
Taking the table above and my formula, d=10m(b-1)2 which was used for multiples in squares. I tried d=10m(l-1)(w-1) to obtain a general formula for consecutive multiples of the rectangle.
Two by six multiples of five
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10 35 Difference
x45 x20 700
450 700 450
250
Using the calculations above, I am going to try and prove that my formula d=10m(l-1)(w-1), is accurate in calculating consecutive multiples in the rectangle.
d=10m(l-1)(w-1)
d=(10x5)(1x5)
d=50x5
d=250
X
X+25
X+10
X+35
X(X+35) (X+10)( X+25) Difference
=X2+35X =X2+25X+250+10X X2+35X+250
=X2+35X+250 - X2+35X
=250
My formula is correct as the answer I got is two hundred and fifty, I also got that same answer in my number term working, and also in my algebra. I can therefore be confident that this formula works, and can use it as a general formula.
From this project I have seen how interesting it is to investigate a nu,berg id and work out different formulas.
All the calculations that I have done in my project are correct.
The formulas that I worked out are also correct and may be used as general formulas.
Eli rose maths course-work
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