GCSE: Number Stairs, Grids and Sequences
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Number Grid
(Could try another example here but the difference will be 10) Move to the general case to see if the answer for a 2 x 2 grid is always 10 (general case means true for any 2 x 2 square on this number grid) n n+1 n+10 n+11 n means starting any where in the grid n+1 because it is one more further on n+10 because each row goes up 10 at a time n+11 because this is one more than the previous number So multiplying as before we get; Top left hand number x Bottom right hand number n x (n+11)
 Word count: 2120

investigate how many winning lines there are in a 7x9 grid.
I am going to put my results in a table to enable me to spot any patterns that occur. Winning Lines n (grid size) Horizontal Vertical Diagonal 4 4 4 2 5 10 10 8 6 18 18 18 As we can see horizontal and vertical winning lines are the same in square grids. Horizontal and Vertical Rules The first difference is not a constant, therefore we do not have a linear sequence. The fact that the second difference is 2, suggests that the rule must contain n�. Rule: n(n3) I predict that in grid size 7x7, the horizontal and vertical number of winning lines will be 28.
 Word count: 1303

Investigate different shapes in different sized number grids.
88 89 90 91 92 93 94 95 96 97 98 99 100 From the four shapes I am using I will do different sizes of each in my paper, the sizes used are as follows: Square 2x2 3x3 4x4 5x5 Once I have done this I will work out the formula and work out what 6x6 would be then do 6x6 to show that it is correct. NOTE: the 6x6 test will only apply for the 10x10 grid, for the 5x5 grid I will do up to 4x4 then work out the formula and find out what 5x5 will be.
 Word count: 6562

Number Grid
I will now assign the first number in the box with x, thus meaning that the number next to it will be x+1 and the one directly below x will be x+10 and then one next to that will be x+11. x x+1 x+10 x+11 This box shows how the numbers layout in each 2 x 2 box. I will also represent the difference of ad and bc with the letter y. Therefore, the two variables will be x and y: x = Independent variable y = Dependant Variable I know: y = bc  ad Result: y = 10 Proof: y = (x+1)
 Word count: 6216

Number Grids
3 x 3 Grids I will now investigate 3x3 grids 5 6 7 15 16 17 25 26 27 7 x 25 = 175 5 x 27 = 135 175  135 = 40 12 13 14 22 23 24 32 33 34 14 x 32 = 448 12 x 34 = 408 448  408 = 40 48 49 50 58 59 60 68 69 70 50 x 68 = 3400 48 x 70 = 3360 3400  3360 = 40 71 72 73 81 82 83 91 92 93 73 x 91 = 6643 71 x 93 =
 Word count: 3576

My coursework task is to investigate why, in a number grid square of 1100, when a section of two by two squares is extracted and the two opposite squares are multiplied and then subtracted the result is always 10.
Now that I have proven the formula with 2x2 squares extracted from a 1100 grid square, will the formula still apply to a 3x3 square and will the result be 10? (3x21) (1x23) = 40 (36x24) (34x56) = 40 (67x89) (69x87) = 40 (80x98)(78x100)=40 As we can see the result is clearly 40, no matter what section of the number square we use. We can once again show why the result is always 40 by using algebra as formulas. (P+20) (P+2)  P (P+22) Now we can put the formula to the test by using it with numbers: (57x75) (55x77)
 Word count: 2571

Number grid
(2x2) = (2 rows x 2 columns) 4 numbers (3x3) = (3 rows x 3 columns) 9 numbers (4x4) = (4 rows x 4 columns) 16 numbers I will begin my investigation by randomly selecting two (2x2) number grids. I will then multiply the top left by the bottom right of the (2x2) number grids and then multiply the top right by the bottom left, after that I will subtract the two totals to find the difference. if there is any pattern then I will randomly select a third number grid and try to predict the result. I will do the same thing again but instead of using a (2x2)
 Word count: 681

Number Grid
We will then see if the difference is 10, which I predict it will be. 57 x 68 = 3876 58 x 67 = 3886 As I predicted this gives a difference of 10. This means there is a general rule for all the boxes in this method. To find this rule I will have to use the algebra in the boxes as shown above. n (n +11) = n? + 11n We will need to find out the algebra product of the opposite way around too. (n+10)(n+1) = n? +n+10n +10 = n? +11n +10 This proves that the differences between the two products are 10.
 Word count: 3043

Maths Coursework on Diagonal Differences
a 5 by 5 grid and an 11 by 11 grid, and find the formula to see if they are the same for a 10 by 10 grid. I will also do an extension by doing a rectangle as well as a square and then find the diagonal difference and the formula for this. I am going to find the formula by finding the diagonal difference of all the sizes within the 10 by 10 grid, and then try to find any pattern, which would help me in finding the formula by drawing a grid.
 Word count: 2332

Number stairs
the 3step stair I can start to establish if there is a pattern I need to find a pattern so that I can find an algebra formula to represent this pattern and use the formula for the 10 by 10 Number Grid By looking at the 3 step stair diagram we know that there are 6 boxes and I will assume in a 3 step stair the bottom left box is equal to X, therefore in my 3step stair x=71 Now I am going to use the values in algebra equations and below and this is how it is going
 Word count: 4419

Number Grid
Prediction I predict that square number 5 will have a difference of 10. 5x16 = 80 6x15 = 90 Difference = 9080 = 10 Square Number Difference 5 10 I will now try this with an 8x8 grid, to help with finding the formula for any size grid. Grid Two 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
 Word count: 2816

Find relationships between the stair total and the position of the stair shape on the grid for a three step stair.
+ (x + 2) + (x + 10) + (x + 11) + (x + 20) Total is T = (xR+c) where R is the reference number and C is the coefficient Therefore gathering the terms I found the formula to be 6x + 44 Other Step Stairs on A 10X10 Grid By knowing this now I then went onto work out other steps on a 10x10 grid. I started by working out a 2 step stair using the method I used to work out the 3 step stair. 11 1 2 x + (x + 1)
 Word count: 2296

My aims throughout this investigation are for each step stair is to investigate the relationship between the stairs and the position of the step stair.
I got this by adding all the numbers within the three step stair (the bold red numbers). The letter 'N' represents the position of the three step stair. In this case 'N=1'. The reason for this is because the step stair starts with the number 1. The letter 'T' represents the total of the step stair. The total for this step stair is 50. These are the expressions I will be using through out the investigation in reference with what it stands for. So far the results have indicated no pattern so I will change the position the step stair.
 Word count: 4075

"Multiply the figures in opposite corners of the square and find the difference between the two products. Try this for more 2 by 2 squares what do you notice?"
To test this I will move the square along a place and see what the difference is. 2 3 12 13 3 x 12 = 36 2 x 13 = 26 This is interesting because the difference seems to stay constant. This is possibly because I have only moved it across the grid. I will now randomly take a square from the bottom right hand corner of the grid and see if the difference stays at 10.I have used a random number generator from 1  89 located on my computer to make sure the random method of choosing the placement of the square remains unbiased.
 Word count: 3344

"Multiply the figures in opposite corners of the square and find the difference between the two products. Try this for more 2 by 2 squares what do you notice?" Investigate!
To test this I will move the square along a place and see what the difference is. 2 3 12 13 3 x 12 = 36 2 x 13 = 26 This is interesting because the difference seems to stay constant. This is possibly because I have only moved it across the grid. I will now randomly take a square from the bottom right hand corner of the grid and see if the difference stays at 10.I have used a random number generator from 1  89 located on my computer to make sure the random method of choosing the placement of the square remains unbiased.
 Word count: 4680

Mathematics Layers Coursework
Algebra: The formula for this problem is: n This is were n equals the size of the grid. When we substitute numbers in it is: 6 Part Two A For part two A I will investigate the relationship between number of arrangements and the size of the grid when there are two layers of cubes on a grid of six squares. Method: I am, like part 1, going to draw out all of the combinations for the specifications above to find out how many combinations there are.
 Word count: 2679

Number Grids
(4x4) (5x5) (6x6) up to (10x10) I repeated these in different areas. Here are my results: (2x2) every time gave a difference of 10 (It was 10 but as the question asked for a difference I changed this to a positive consistently) (3x3) every time gave a difference of 40 (4x4) every time gave a difference of 90 (5x5) every time gave a difference of 160 (6x6) every time gave a difference of 250 (7x7) every time gave a difference of 360 (8x8) every time gave a difference of 490 (9x9) every time gave a difference of 640 (10x10)
 Word count: 1095

Investigate the relationship between different stairs placed in different places in a number square, which is numbered from one to one hundred.
This also means that every time you move the twostep stair one to the right the product will go up by three. 23 23+13+14=50 13 14 33 33+23+24=80 23 24 43 43+33+34=110 33 34 I have found out that when you move the stairs up by one the product of the twostep stair goes up by 30. This also means that when you move the twostep stair down by one the product will go down by 30. I have also found out the nth term for all twostep stairs that 3n+11 will equal the sum off all the numbers in the stair.
 Word count: 1631

Number Grid
General Rule For 3x3 If we look at the 3x3 square algebraically it would look like this: Top Left x Bottom Right (X)(X+22)=X2+22X Top Right x Bottom Left (X+2)(X+20)=X2+22X+40 Prove X=2 Top Left x Bottom Right (2)2+22(2)=4+44=48 Top Right x Bottom Left (2)2+22(2)+40=2+44+40=88 8848=40 QED This proves the difference is always 40 4x4 I will now be looking at a 4x4 square to see if it follows a pattern Example 1 Top Left x Bottom Right 32 x 65= 2080 Top Right x Bottom Left 35 x 62= 2170 21702080=90 The difference is 90.
 Word count: 4251

Number grid
3 x 3 square the difference will always be 40 Algebra I will assign a letter to the first number in the 3x3square, n. The right hand top corner will therefore be n+2 The left hand bottom corner will then be n+20 The corner diagonally across from it will be n+22 I will then multiply the corner numbers, as shown in the above example. Top Left hand corner x bottom right hand corner = n(n+22) = n? + 22n Top right hand corner x bottom left hand corner = (n+20)(n+2)
 Word count: 1447

Investigation of diagonal difference.
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 22 x 13 = 286 => 23 x 12 = 276 => 286 276 = 10 The diagonal difference is 10 Is the diagonal difference of a 2x2 cutout the same anywhere on the 10x10 grid?
 Word count: 7972

The Open Box Investigation
If the volume increases then I will make the cut size smaller or larger by 0.1cm again. I will repeat this for grid sizes of 18cm x 18cm and 24cm x 24cm. I have found the following results: Grid Size: 12cm x 12cm Cut Size Calculation Volume 1cm 10 x 10 x 1 100cm3 2cm 8 x 8 x 2 128cm3 3cm 6 x 6 x 3 108cm3 1.9cm 8.2 x 8.2 x 1.9 127.8cm3 2.1cm 7.8 x 7.8 x 2.1 127.8cm3 Grid Size: 18cm x 18cm Cut Size Calculation Volume 1cm 16 x 16 x 1 256cm3 2cm 14 x 14 x 2 392cm3 3cm 12 x 12 x 3 432cm3 4cm 10 x 10 x
 Word count: 1034

Number Grid
that the difference is always 10, so anywhere you put a 2x2 window on a 10x10 grid you will always get the same difference of 10. This can also be shown algebraically. n n+1 n+10 n+11 We can change this equation into: (n+1)(n+10)n(n+11) When multiplied outside of the brackets: n�+ 10n + n + 10  n�11n This can be simplified to: n� + 11n + 10  n�  11n = 10 The n� and 11n cancel each other out therefore you are left with 10.
 Word count: 5603

Number grids
Now I am going to choose a general position of my square, the top right will be "M". Mx(M+11)=M2+11M (M+1)x(M+10)= M2+11M+10 M2+11M+10  M2+11M=10 This has proven that any two by two squares on this number grid, I choose and take the products away the answer will always be 10. Now I am going to investigate when we have a 3 by 3 box in our grid. 14x36=504 16x34=544 544504=40 45x67=3015 47x65=3055 30553015=40 The answer is 40 each time I have done it. There looks as though there is a pattern going on here the answer will always be 40.
 Word count: 1600

3 step stair investigation
8 > 6N + 36 7 by 7 > 6N + 32 6 by 6 > 6N + 28 91 92 93 94 95 96 97 98 99 100 81 82 83 84 85 86 87 88 89 90 71 72 73 74 75 76 77 78 79 80 61 62 63 64 65 66 67 68 69 70 51 52 53 54 55 56 57 58 59 60 41 42 43 44 45 46 47 48 49 50 31 32 33 34 35 36 37 38 39 40 21 22 23 24 25 26 27 28 29 30 11
 Word count: 1849