# Investigate the relationships between the numbers in the crosses.

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Introduction

GCSE Maths Coursework 2 – Assessment A01

Aim

- To investigate the relationships between the numbers in the crosses.

Hypotheses

- I predict that several connections will be found, which will help relate to crosses in a 100 square grid.
- I predict that were ever the cross is on the grid that the result will always be the same, for each cross.
- I predict that when the cross is changed, that the original equation will still apply but with a division number.
- That knowing one number on the cross that all others are able to be found by adding or subtracting a fixed number.

Original cross

36 74 a

- 46 47 83 84 85 d x b

56 94 c

Rule:

- (Left x right number) – (top x bottom number) = 99

- This rule states that if a cross is situated anywhere on a grid, then the left number multiplied by the right number subtract the top number multiplied by the bottom number always gives the number 99.

Proof:

- (45 x 47) – (36 x 56) = 99
- (87 x 85) – (74 x 94) = 99

Equation:

- (d x b) – (a x c) = 99

- This algebraic equation shows that were ever the cross is situated on the grid that the put come of the left number multiplied by the right number subtract the top number multiplied by the bottom number always equals 99.

Finding numbers:

36 (a) 36 (b-11)

(a+9) 45 46 47 (a+11) (b-2) 45 46 47 (b)

(a+10) 56 (a+20) (b-1) 56 (b+9)

36 (c-20) 36 (l-9)

(c-11) 45 46 47 (c-9) (l) 45 46 47 (l+2)

(c-10) 56 (c) (l+1) 56 (l+11)

36 (x-10)

(x-1) 45 46 47 (x+2)

(x) 56 (x+10)

Finding numbers in equations:

- If a is known:

= (a+9)(a+11) – (a)(a-10)

= (a +11a+9a+99) – (a +20a)

= (a +20a+99) – (a +20a)

= 99

- If b is known:

= (b)(b-2) – (b+9)(b-11)

= (b -2b)

Middle

- This rule states that if the top left number add the bottom left number multiplied by the top right number add the bottom right number, subtract the top left plus the top right multiplied by the bottom left plus the bottom right number. With number, it’s divided by 4 to give 99.

Proof:

- (35 + 55)(37 + 57) – (35 + 37)(55 + 57) = 99

4

- (73 + 93)(75 + 95) – (73 +75)(93 + 95) = 99

4

Equation:

- (a + d)(b + c) – (a + b)(d + c) = 99

4

- This algebraic equation also proves that were ever this particular cross is situated on the grid, that the out come of the top left number add the bottom left number multiplied by the top right number add the bottom right number, subtract the top left plus the top right multiplied by the bottom left plus the bottom right number, then that total number divided by four always gives 99

Symmetrical cross

26 54 a1

36 64 a2

44 45 46 47 48 72 73 74 75 76 d1 d2 x b2 b1

56 84 c2

66 94 c1

Rule:

- (1st left number x end right number) – (top number – bottom number) = 396

- This rule states if the 1st left number multiplied by the end right number, subtract the top number multiplied by bottom number the answer is always 396.

Proof:

- (44 x 48) – (36 x 66) = 396

- (72 x 76) – (54 – 94) = 396

Equation:

- (d1 x b1) – (a1 x c1) = 396

- This algebraic equation shows were ever this particular cross is situated on the grid that if the 1st left number multiplied by the end right number, subtracted by the top number multiplied by the bottom number always gives 396.

Finding numbers:

26 (a) 26 (b-22)

36 36

(a+18) 44 45 46 47 48 (a+22) (b-4) 44 45 46 47 48 (b)

(a+20) 56 (b-2) 56

66 (a+40) 66 (b+18)

26 (c-40) 26 (d-18)

36 36

(c-22) 44 45 46 47 48 (c-18) (d) 44 45 46 47 48

(c-20) 56 (d+2) 56

66 (c) 66 (d+22)

26 (x-20)

36

(x-2) 44 45 46 47 48

(x) 56

66 (x +20)

Finding numbers in equations:

- If a is known:

= (a+18)(a+22) – (a)(a+40)

= (a +22a+18a+396) – (a +40a)

= (a +40a+396) – (a +40)

= 396

- If b is known:

= (b)(b-4) – (b-22)(b+18)

= (b -4b) – (b +18b-22b-396)

= (b -4b) – (b -4b-396)

= -396

- If c is known:

= (c-22)(c-18) – (c)(c-40)

= (c -18c-22c+396) – (c -40c)

= (c -40c+396) – (c -40c)

= 396

- If d is known:

= (d)(d+4) – (d-18)(d+22)

= (d +4d) – (d +22d-18d-396)

= (d +4d) – (d +4d-396)

= -396

- If x is known:

= (x-2)(x+2) – (x+20)(x-20)

= (x +2x-2x-4) – (x -20x+20x-400)

= (x +x-4) – (x +x-400)

= 396

- This also implies that the sum (d1 x b1) – (a1 x c1) = 396 can come out positive or negative, pending on the number used to solve the equation.

- This equation also relates back to the first equation of (d x b) – (a x c) = 99 as 369 is a multiple of 99 as it equals 99 multiplied by 4 equals 396. This also shows that the cross is proportional, as the length of the cross has increased by 2, 2 multiplied by 2 equals 4.

Proof

- (44 x 48) – (36 x 66) = 99

4

- (72 x 76) – (54 – 94) = 99

4

Equation

- (d1 x b1) – (a1 x c1) = 99

4

- This supports the new rule, which links the crosses together to shows the relationship of numbers in the cross equalling 99.

- If the cross had a length of three numbers, the rule would still work.

Proof

- (43 x 49) – (16 x 76) = 891

- 891 = 99

9

- (61 x 67) – (44 x 94) = 891

- 891 = 99

9

Equation

- (b1 x d1) – (a1 x c1) = 99

Conclusion

- Example: (11.5 x 12.5) – (2 x 22) = 99.75

Cross with no same length

- I found that if the original cross was developed to have no two sides the same that there wouldn’t be a constant number able to be found, and that it wouldn’t relate at all back to the original cross.

Conclusion

- From this coursework, I found that (Left x right number) – (top x bottom number) = 99, with changes, can apply to any and every cross, made on the 10 x 10 grid of squares. I have also concluded and proved my hypnosis’s, and found a conclusion to all. I have also extended my work to find solutions to finding the number 99 in different ways, this has also proved my first equation, and supports and proves the equation. I also found that if the cross had two or more sides the same, that it was possible to relate back to the original cross or equation, pending on whether the cross developed length ways or width ways.

Evaluation

- If I did this coursework again I wouldn’t change the way I set out my work, or disclose any of the work I have done already. I would do more research into grids, to see if this at all effected the cross, and find different rules that apply to all crosses through out. I would also do the cross more extended i.e. increasing in one length side and one width side, to see if the equation could apply to more, and that if the original cross could relate to any cross made on the 10 x 10 square grid.

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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