Mechanics Maths Coursework On Cubes

Number Squares

In this coursework I will be investigating any patterns or correlations that occur within a set scenario. I will conduct this experiment using 10x10 number squares.

1 2 3 4 5 6 7 8 9 10

11 12 13 14 15 16 17 18 19 20

21 22 23 24 25 26 27 28 29 30

31 32 33 34 35 36 37 38 39 40

41 42 43 44 45 46 47 48 49 50

51 52 53 54 55 56 57 58 59 60

61 62 63 64 65 66 67 68 69 70

71 72 73 74 75 76 77 78 79 80

81 82 83 84 85 86 87 88 89 90

91 92 93 94 95 96 97 98 99 100

I then will draw a 2x2 grid anywhere on the larger grid. This will result in us selecting four numbers. I will refer to the smallest number in the top left as the starting number. For Example if our starting number was 47, we would have a 2x2 grid which would resemble:

47

48

57 58

As you can see, this will result in us having four numbers. Next we multiply the number in the top left with the number in the bottom right, and do the same with the top right and bottom left numbers.

This procedure leaves us with two results: 47x58=2726

57x48=2736

Finally we subtract the results from each other which give us the difference between them.

2736-2726=10

As we can see this first answer leaves us with 10. I will now conduct this procedure with as many different starting numbers as needed for some kind of pattern to emerge.

Next, we will start at the beginning with the number 1 as the starting number. Remember this means that this number will be the top left figure. On a 10x10 grid, a 2x2 box with the number one in the top left corner would look like:

1

2

11 12

Now we carry out our equations: 1x12=12

2x11=22

It is now obvious that we have once more gained a difference of 10.

We will now investigate the algebra behind this equation to see if there is a real mathematical reason for this occurrence.

N

N+1

N+10 N+11

I have used the letter N to represent the starting number which occurs in the top left of the grid. I have then labelled all the other numbers in the grid in relation to the starting number, as can been seen on the grid.

I will now carry out my equations using an algebraic formula instead of actual numbers. This results in: N(N+11)=NÐ'І+11N

(N+1)(N+10)=NÐ'І+11N+10

Difference=10

The fact that the formula are identical with exception to the +10 at the end of the second equation proves beyond all doubt that in the situation of the top left of a 2x2 grid being multiplied with the bottom right and then calculating the difference between the product and that of the top right and bottom left will always result in 10.

However, there are certain values that cannot be N, as attempting the process with them does not work. This is due to the fact that some numbers will result in the smaller grid no longer fitting on the larger grid, rendering our formula incomplete. For example if N was 91, the bottom two squares would extend beyond the 10x10 grids boundaries.

1 2 3 4 5 6 7 8 9 10

11 12 13 14 15 16 17 18 19 20

21 22 23 24 25 26 27 28 29 30

31 32 33 34 35 36 37 38 39 40

41 42 43 44 45 46 47 48 49 50

51 52 53 54 55 56 57 58 59 60

61 62 63 64 65 66 67 68 69 70

71 72 73 74 75 76 77 78 79 80

81 82 83 84 85 86 87 88 89 90

91

92 93 94 95 96 97 98 99 100

As you can see we are missing the values needed with which to complete our multiplications! Therefore we know that the starting number must be positioned at least the second grid's size away from the bottom or right edges.

If we disregard the values which extend the grid beyond the 10x10 square, we are left with a formula which will work with all remaining values. I therefore predict in my next equation with the starting number of 87, that when I deduct the two totals I will be left with ten.

Therefore if N=87 87Ð'І+957= 8526

87Ð'І+957+10= 8536

87

88

97 98

87x98= 8526

97x88= 8536

This check shows that my prediction was correct.

Now that I know this about 2x2 grids I will now investigate the effect a 3x3 grid has on the same procedure.

A 3x3 grid imposed on a 10x10 grid with a starting number of 1 will look like the following:

1 2 3

11 12

13

21 22 23

Therefore the equation goes as follows: 1x23=23

3x21=63

We can now see that there is a difference of 40 between the two totals. We cannot be sure at this point that this will be consistent with