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• Level: GCSE
• Subject: Maths
• Word count: 2057

# Number grids

Extracts from this document...

Introduction

Number grids

My task is to find an algebraic rule for different sized squares in a set sized number grid.

To do this I will establish my algebraic rule by creating a 10×10 square and marking out 3 different sized squares inside this square. I will then work out the rules for these individual squares and combine them to create my overall rule.

 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 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

I have marked out my smaller squares inside the grid and will now work out an algebraic rule:

To find my algebraic rule I will times the opposite corners in the inset squares and take the numbers away from each other to find the difference.

2×2-

55×66=3630

56×65=3640

3640-3630 = 10

89×100=8900

90×99=8910

8910-8900=10

22×33=726

23×32=736

736-726=10

27×38=1026

28×37=1036

1036-1026=10

After multiplying the corners of the 2×2 squares I then took the lowest away from the highest. This number is always 10.

In this section: a represents the number in the top left hand corner of the inset square.

 a a+1 a+10 a+11

(a+1)×(a+10)-a×(a+11)Here I have multiplied the opposite corners of the grid

[a²+11a+10]-[a²+11a] Here I have multiplied out the brackets and simplified the rule

a²+11a+10    -

a²+11a _        Here I have subtracted the two sections to prove my overall rule.

0 +  0 +10

=10

Middle

10×1²=10

The number in the 2×2 squares was 10 so this proves my formula correct.

To extend my work I will create a different sized number grid and calculate a formula that will work for the difference in any sized square in any sized grid.

 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

As in my original grid I have marked out individual squares and will now calculate the formula for them

2×2

1×9=9

2×8=16

16-9=7

6×14=84

7×13=91

91-84=7

25×33=825

26×32=832

832-825=7

36×44=1584

37×43=1591

1591-1584=7

41×49=2009

42×48=2016

2016-2009=7

 a a+1 a+7 a+8

a×(a+8)-(a+1)×(a+7)

[a²+8a+7]-[a²+8a]

a²+8a+7  -

a²+8a

0 +0 +7

=7

3×3

1×17=17

3×15=45

45-17=28

10×26=260

12×24=288

288-260=28

29×45=1305

31×43=1333

1333-1305=28

33×49=1617

35×47=1645

1645-1617=28

5×21=105

7×19=133

133-105=28

 a a+1 a+2 a+7 a+8 a+9 a+14 a+15 a+16

a×(a+16)-(a+2) ×(a+14)

[a²+16a+28]-[a²+16a]

a²+16a+28   -

a²+16a

0 + 0+28

=28

4×4

1×25=25

4×22=88

88-25=63

22×46=1012

25×43=1075

1075-1012=63

25×49=1225

28×46=1288

1288-1225=63

 a a+1 a+2 a+3 a+7 a+8 a+9 a+10 a+14 a+15 a+16 a+17 a+21 a+22 a+23 a+24

a×(a+24)-(a+3)×(a+21)

[a²+24a+63]-[a²+24a]

a2+24a+63 -

a²+24a

0+0+63

=63

Before creating an overall formula I will create a formula for my 7×7 grid. This will allow me to create my overall formula:

a×(a+8(n-1)) – ((a+(n-1)×(a+7(n-1)))

This cancels down to

(a²+8a(n-1))-(a²+8a(n-1)+7(n-1)²)

To work out my overall formula I will subtract the first part of the Formula from the other:

Conclusion

a2+aB(A-1)+a(Z-1)+(Z-1)B(A-1)

a2+aB(A-1)+a(Z-1) –

0 +     0    +    0   +(Z-1)B(A-1)

=B(Z-1)(A-1)

=15(Z-1)(A-1)

Looking at this formula and my previous work with squared I can see that the general formula for the difference could be:

B(Z-1)(A-1)

I will now prove this by creating a 7×5 grid and using the same process before, make up a general rule for any sized rectangle in any sized grid.

 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

2×3

1×16=16

2×15=30

30-16=14

18×33=594

19×32=608

608-594=14

 a a+1 a+2 a+7 a+8 a+9

a×(a+9) – (a+2)×(a+7)

a2+9a – a2+9a+14

a2+9a+14

a2+9a –

0+0+14

=14

6×2

15×27=405

22×20=440

440-405=35

22×34=748

29×27=783

783-748=35

 a a+1 a+2 a+3 a+4 a+5 a+7 a+8 a+9 a+10 a+11 a+12

a×(a+12) – (a+5)×(a+7)

a2+12a – a2+12a+35

a2+12a+35

a2+12a –

0+0+35

=14

I will now create my overall formula for any sized rectangle in any sized grid.

The letters represent the same as previously.

a×(a+B(A-1)+(Z-1)) – a+(Z-1)) × (a+B(A-1))

a2+aB(A-1)×a(Z-1)+(Z-1)B(A-1)

a2+aB(A-1)×+a(Z-1) –

0 +     0    +    0   +(Z-1)B(A-1)

=B(Z-1)(A-1)

Therefore I conclude that my overall formula for the difference is B(Z-1)(A-1). I will check this by inserting numbers into my formula.

B=7

Z=6

A=2

B(Z-1)(A-1)

7×(6-1)(2-1)

=7×(5×1)

=7×5=35

This proves my formula for the difference correct as in the 6×2 grids the difference was always 35.

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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