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Introduction

GCSE Maths

Number Stairs

Part 1:For the other 3-step stairs, investigate the relationship between the stair total and the position of the stair shape on the grid.

Here we have a 10 by 10 grid. In the grid, there is a shape called the 3-step stair

 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 12 13 14 15 16 17 18 19 20 1 2 3 4 5 6 7 8 9 10

The number at the bottom left of the 3-step stair is 25. This is called the Step-number.

The sum of all the numbers in the 3-step stair is 194.

25 + 26 + 27 + 35 + 36 + 45 = 194

This is called the Step-total.

I am going to investigate the relationship between the stair total and the position of the stair shape on the grid using the first step-number on the grid.

 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 12 13 14 15 16 17 18 19 20 1 2 3 4 5 6 7 8 9 10

The first 3-step stair is made up of

1 + 2 + 3 + 11 + 12 + 21 = 50

If I move the 3-step stair 1 unit to the right, the 3-step stair would be made up of

2 + 3 + 4 + 12 + 13 + 22 = 56

If I move the 3-step stair another unit to the right, the 3-step stair would be made up of

3 + 4 + 5 + 13 + 14 + 23 = 62

I have created a table of a few results of the 3-step stairs

 Numbers in 3-step stair Step-number Step-total 1, 2, 3, 11, 12, 21 1 50 2, 3, 4, 12, 13, 22 2 56 3, 4, 5, 13, 14, 23 3 62 4, 5, 6, 14, 15, 24 4 68 5, 6, 7, 15, 16, 25 5 74 6, 7, 8, 16, 17, 26 6 80

From this information, you can see that;

1. Each 3-step stair can be divided evenly by 2
2. Each step-total increases by 6 starting from 50
3. The step-number goes up by 1 and the step-total goes up by 6.

Middle

43

44

45

46

47

48

49

50

31

32

33

34

35

36

37

38

39

40

n + 20

22

23

24

25

26

27

28

29

30

n + 10

n + 11

13

14

15

16

17

18

19

20

n

n + 1

n + 2

4

5

6

7

8

9

10

Since all the numbers in the stair-shape adds up to the step-total, if I simplify all the values in the 3-step stair, I should get a formula for the step-total

n + (n + 1) + (n + 2) + (n + 10) + (n + 11) + (n + 20)        =        6n + 44

Therefore, the step-total should equal to 6n + 44

I called step-total t

For stair-step 1, n = 1

t = 6n + 44

= 6(1) + 44

= 6 + 44

= 50

I have proven that the formula works for the first 3-step stair. To verify that this formula works for every other 3-step stair, I will test the first example given, stair-step 25. The step-total should equal 194.

t = 6n + 44

= 6(25) + 44

= 150 + 44

= 194

The formula works for all the step-numbers on the 10 by 10 grid. The relationship between the stair-totals and the position of the stair-shape on the grid is that if you multiply the step-number by 6 and the add 44, you should find the step-total.

Part 2:Investigate further the relationship between the stair totals and other step stairs on other number grids.

I have worked out the formula for the 10 by 10 grid to be t = 6n +44 but this formula only works for a 10 by 10 grid. I will investigate the relationship between the step-total, step-number and the grid size.

In the 10 by 10 grid I noticed that the numbers 1, 11 & 21 etc increases by 10. If you subtracted n from the number above n, you would get the grid size number Therefore, the step-total and the step-number relate with the grid size.

 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 12 13 14 15 16 17 18 19 20 1 2 3 4 5 6 7 8 9 10

In this example, n is 25 and the number above n is 35. If I subtract n from 35, I would get 10. This is the grid size number.

The relationship between the step-number and the rest of the numbers are that;

The number 35 has a difference of 10 from the step-number. The number 45 has a different of 20 from the step-number, which is 2 multiplied by 10. The number 26 has a difference of 1 from the step number, which is 1 multiplied by 10, minus 9.The number 27 has a difference of 2 from the step number, which is 1 multiplied by 10, minus 8.

The number 36 has a difference of 11 from the step number, which is 1 multiplied by 10, plus 1

With this information, I will create a formula for each of the numbers in the 3 step-stair shape. The Grid size will be called g.

 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 n+2g 46 47 48 49 50 31 32 33 34 n+g n+(g+1) 37 38 39 40 21 22 23 24 n n+(g-9) n+(g-8) 28 29 30 11 12 13 14 15 16 17 18 19 20 1 2 3 4 5 6 7 8 9 10
 n= 25 n + ( g – 9)= 25 + (10 – 9)= 25 + 1=26 n + ( g – 8 )= 25 + (10 – 8)= 25 + 2=27 n + g= 25 + 10= 35 n + ( g + 1 )= 25 + (10 + 1)= 25 + 11=36 n + 2g= 25 + (10 x2)= 25 + 20=45

Conclusion

n+6g – 14

From this information, I have come up with another equation that will work with a grid of any size;

6n + 6g – ( 16 ± 2d )

Where d is the difference of the grid from a 10 by 10 grid

I will test this new formula with the 10 by 10 grid and the 9 by 9 grid.

Step-number = 1

Grid size = 10 by 10

t = 6n + 6g – ( 16 ± 2d )

= 6(1) + (6 x 10) – [16  ± (2 x 0)]

= 66 – 16

= 50

The new equation has worked for the 10 by 10 grid

Step-number = 1

Grid size = 9 by 9

t = 6n + 6g – ( 16 ± 2d )

= 6(1) + (6 x 9) – [16 – (2 x 1)]

= 60 – 14

= 46

The new equation has also worked for the 9 by 9 grid

This formula should even work for a smaller scale, such as a 5 by 5 scale

 21 22 23 24 25 16 17 18 19 20 11 12 13 14 15 6 7 8 9 10 1 2 3 4 5

The answer we should get is;

1 + 2 + 3 + 6 + 7 + 11 = 30

t = 6n + 6g – ( 16 ± 2d )

= 6(1) + (6 x 5) – [16 – (2 x 5)]

= 36 – 6

= 30

I have proven that my formula works with any grid

Conclusion:

Here I have put all the formula I have come up with. This formula will apply to a grid of any size;

t = 6n + 6g – ( 16 ± 2d )

Where;

t   : step- total

n  : step-number

g  : grid size

d  : difference of grid size from a 10 by 10 grid

In this project I have found out many ways in which to solve the problem I have with the stair-shape being in various positions with different sizes of grids. The way I have made the calculations less difficult is by creating a main formula that works for all the different circumstances.

This student written piece of work is one of many that can be found in our GCSE Number Stairs, Grids and Sequences section.

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