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• Level: GCSE
• Subject: Maths
• Word count: 1467

# Number Stairs

Extracts from this document...

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

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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