# Number Stairs

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Introduction

Number Stairs

Mathematics GCSE Coursework

Part 1

Introduction

I am going to investigate relationships between 3 step stairs on a ten by ten grid. I also intend to find algebraic expressions to link the 3 step stairs and there positions and totals.

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 |

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

The total of the 3 the step stair is the sum of all the numbers included in it.

Excluded

The numbers in the grid with a lighter colour do not have totals because a 3 step stair can not be formed with them.

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 |

Expressions

I am going to use these letters to show certain numbers:

N= bottom left hand number on the step stair, the position (the yellow number above)

T= total of the step stair

T(t-1)= total of step stair before

G= the grid size

R= horizontal row total

R(r-1)= total of horizontal row before

C= vertical column total

C(c-1)= total of column before

The Task

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

Data Collection

I am going to display some totals and positions in the grid.

The value before the equals sign is the bottom left hand position in the step stair, also known as N.

1=50, 2=56, 3=62, 4=68, 5=74, 6=80, 7=86 and 8=92. I can’t do 9 or 10 because there are not enough stairs to complete the step stair grid.

I noticed that the difference between all the totals is 6:

1=50

6

2=56

6

3=62

6

4=68

6

5=74

6

6=80

6

7=86

6

8=92

I am now going to investigate the second row,

11=110, 12=116, 13=122, 14=128, 15=134, 16=140, 17=146, 18=152. These also have the same difference between the numbers i.e. 6.

This would mean the formula T=T(t-1)+6 would work for the second row as well as the top row, apart from N=19 and 20.

Middle

T=128+6

T=135 where N=15

I checked my answer was correct by adding up the step stair N=15,

15+16+17+25+26+35=135

This shows the formula works for N=15

This shows that each total on the horizontal row is 6 more than the previous total, except the totals in the 1, 9 and 10 vertical row and 9 and 10 horizontal row. The formula to show this is T=T(t-1)+6.

I am now proceeding to investigate vertical columns; I am looking at the totals ad comparing them as they proceed up the row.

1=50

60

11=110

60

21=170

60

31=230

60

41=290

60

51=350

60

61=410

60

71=470

I could do not do 81 or 91 because using either as the numbers as N I could not complete the 3 step stair.

I am now going to investigate the second column,

2=56, 12=116, 22=176, 32=236, 42=296, 52=356, 62=416, 72=476

I could immediately see from this that the difference between the totals is 60.

I am now going to investigate the third column,

3=62, 13=122, 23=182, 33=242, 43=302, 53=362, 63=422, 73=482

I am now going to investigate the fourth column,

4=68, 14=128, 24=188, 34=248, 44=308, 54=368, 64=428, 74=488

I am now investigating the fifth column,

5=74, 15=134, 25=194, 35=254, 45=314, 55=374, 65=434, 75=494

I am now going to investigate the sixth column,

6=80, 26=140, 36=200, 46=260, 56=320, 66=380, 76=440

I am now investigating the seventh column,

7=86, 17=146, 27=206, 37=266, 47=326, 57=386, 67=446, 77=506

I am now investigating the eighth column,

8=92, 18=152, 28=212, 38=272, 48=332, 58=392, 68=452, 78=512

From this I can see that a formula that could arrive at the correct total is

T=T(t-1)+60

I arrived at this formula when I saw that the total can be produced by taking the total before it and adding 60 to it. This formula works on all the numbers except those specified above and the bottom horizontal row because they do not have a preceding total. You can also see that the step stair consists of 6 numbers each of these numbers moves 1 position above themselves becoming greater by 10.

Conclusion

This shows that the formula 6N+28 works for N=7.

I will check my formula again by working out the answer to N=21,

The formula will be,

6x21+28=154

I will check the answer is correct, 21+22+23+28+29+34=154

This shows that the formula works for N=21.

I will now do a final check to see if my formula works by working out the answer to N=15,

The formula will be,

6x15+28=118,

I will now check to see if the answer is correct, 15+16+17+21+22+27=118,

This shows the formula works for N=15,

And the formula 6N+28 works on a 6 by 6 grid.

I am now going to investigate 3 step stairs on a 9 by 9 grid,

73 | 74 | 75 | 76 | 77 | 78 | 79 | 80 | 81 |

64 | 65 | 66 | 67 | 68 | 69 | 70 | 71 | 72 |

55 | 56 | 57 | 58 | 59 | 60 | 61 | 62 | 63 |

46 | 47 | 48 | 49 | 50 | 51 | 52 | 53 | 54 |

37 | 38 | 39 | 40 | 41 | 42 | 43 | 44 | 45 |

28 | 29 | 30 | 31 | 32 | 33 | 34 | 35 | 36 |

19 | 20 | 21 | 22 | 23 | 24 | 25 | 26 | 27 |

10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 |

1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |

Data Collection

The grid below summarises the totals for the step stairs above,

* | * | * | * | * | * | * | * | * |

* | * | * | * | * | * | * | * | * |

370 | 376 | 382 | 388 | 394 | 400 | 406 | * | * |

316 | 322 | 328 | 334 | 340 | 346 | 352 | * | * |

262 | 268 | 274 | 280 | 286 | 292 | 298 | * | * |

208 | 214 | 220 | 226 | 232 | 238 | 244 | * | * |

154 | 160 | 166 | 172 | 178 | 184 | 190 | * | * |

100 | 106 | 112 | 118 | 124 | 130 | 136 | * | * |

46 | 52 | 58 | 64 | 70 | 76 | 82 | * | * |

I can see that the difference between the N numbers as they proceed along the row is still 6, this means the formula T=T(t-1)+6 still applies to the 6 by 6 grid. This can be changed to apply to any step stair on any grid, the new formula is,

T=T(t-1)+S.

This is because each step stair consists of 6 numbers, each of these numbers moves 1 position to the right, this means the step stair before becomes greater by 6 because each of the 6 positions in the step stair move 1 place to the right, 1x6=6 which means the step stair becomes greater by 6. And if the size of the step stair changes the S changes.

But as they proceed up the column the difference has changed again from 36, to 54. From this I can see a recurring pattern that applies to 3 step stairs. The numbers of positions in a step stair multiplied by the grid size, SG, will give the difference between the number above and the number below,

e.g. To find the number f

This student written piece of work is one of many that can be found in our GCSE T-Total section.

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