# Number Stairs

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Introduction

Christopher Little

3N

Number Stairs

Problem:

Look at the stair shape drawn on the 10x 10 grid below.

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 |

This is a 3-step stair.

The total of the numbers inside the stair shape is

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

The stair total for this 3-step stair is 194.

Part 1

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

Part 2

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

I plan to work first on a 10x10 grid. Let the total of the numbers within the stair be S. Let the number at the bottom left of the grid (the ‘stair number’) be n.

On my 10x10 grid, I will first use an s-number of 25, which makes the s-total = 194. I will then work systematically, increasing the s-number by 1 every time. However, the s-number is limited in its placement – it cannot be a multiple of 10, or a multiple of 10 -1 – else it would not fit on the grid. I will record results in the following table:

s-number (n) | 1 2 3 4 5

s-total (S) | 50 56 62 68 74

Middle

n+g+1

n

n+1

n+2

From which, in turn, I can determine the following formula:

S= 6n +4g +4

This is because there are 6 ns in my shape, 4 gs in my shape, and a 4 in my shape.

I will test this with an 11x11 grid, using an s number of 49. These values were chosen arbitrarily.

71 | 72 | 73 |

60 | 61 | 62 |

49 | 50 | 51 |

(6x49) + (4x11) + 4 = 342

49+50+51+60+61+71 = 342

My formula works!

Different Sized Stairs

I now plan on working out a formula which includes the size of the stair. Let h be the stair size. On a 10x10 grid:

2x2 S total = 14 S=3n+g+1

11 | 12 |

1 | 2 |

3x3 S total= 50 S=3n+g+1

21 | 22 | 23 |

11 | 12 | 13 |

1 | 2 | 3 |

4x4 S total = 120 S=3n+g+1

31 | 32 | 33 | 34 |

21 | 22 | 23 | 24 |

11 | 12 | 13 | 14 |

1 | 2 | 3 | 4 |

5x5 S total = 235 S=3n+g+1

41 | 42 | 43 | 44 | 45 |

31 | 32 | 33 | 34 | 35 |

21 | 22 | 23 | 24 | 25 |

11 | 12 | 13 | 14 | 15 |

1 | 2 | 3 | 4 | 5 |

6x6 S total = 406 S=3n+g+1

51 | 52 | 53 | 54 | 55 | 56 |

41 | 42 | 43 | 44 | 45 | 46 |

31 | 32 | 33 | 34 | 35 | 36 |

21 | 22 | 23 | 24 | 25 | 26 |

11 | 12 | 13 | 14 | 15 | 16 |

1 | 2 | 3 | 4 | 5 | 6 |

Length | Formula _

h=1 | S= n +0g +0

h=2 | S= 3n +g +1

h=3 | S= 6n +4g +4

h=4 | S= 10n +10g +10

h=5 | S= 15n +20g +20

h=6 | S= 21n +35g +35

I can see that the coefficient of n is going up in triangle numbers (1, 3, 6, 10…). I can also see that the difference in the constant (and the coefficient of g, as they are identical) is going up in triangle numbers (i.e. 0+1, 1+3, 4+6, 10+10…).

Conclusion

S= h(h+1)h3-hh3-h

2 6 6

I can now test this formula when

n= 25

g= 11

h= 7

7(7+1) x25 + 73-7 x11 + 73-7 = 1372

2 6 6

25+26+27+28+29+30+31+36+37+38+39+

40+41+47+48+49+50+ 51+58+59+60+61+

69+70+71+80+81+91 = 1372

111 | 112 | 113 | 114 | 115 | 116 | 117 | 118 | 119 | 120 | 121 |

100 | 101 | 102 | 103 | 104 | 105 | 106 | 107 | 108 | 109 | 110 |

89 | 90 | 91 | 92 | 93 | 94 | 95 | 96 | 97 | 98 | 99 |

78 | 79 | 80 | 81 | 82 | 83 | 84 | 85 | 86 | 87 | 88 |

67 | 68 | 69 | 70 | 71 | 72 | 73 | 74 | 75 | 76 | 77 |

56 | 57 | 58 | 59 | 60 | 61 | 62 | 63 | 64 | 65 | 66 |

45 | 46 | 47 | 48 | 49 | 50 | 51 | 52 | 53 | 54 | 55 |

34 | 35 | 36 | 37 | 38 | 39 | 40 | 41 | 42 | 43 | 44 |

23 | 24 | 25 | 26 | 27 | 28 | 29 | 30 | 31 | 32 | 33 |

12 | 13 | 14 | 15 | 16 | 17 | 18 | 19 | 20 | 21 | 22 |

1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 |

Notes

The reason the constant is always the same as the coefficient of g is that the same things happen to the both of them, although perpendicular to each other, as illustrated by this diagram:

For every row added on, an equal amount of g and of the constant is added on.

Limitations

There are some limitations to this formula, as there are some limitations to the placement of the staircase. For example, the staircase cannot be larger than the grid, else it wouldn’t all fit. The coefficient of n is also limited, or the staircase wouldn’t entirely fit on the grid. To create a formula to determine where n is allowed to be, I need to introduce a new variable – r for the row that n is on.

I need to eliminate the top few rows as possible locations for n (depending on the size of the staircase):

n< g2- g(h-1)

I also need to eliminate a few columns on the right as possible locations for n (depending on the size of the staircase):

n< rg -(h-1)

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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