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  • Level: GCSE
  • Subject: Maths
  • Word count: 1696

Number Grid Investigation

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Introduction

Number Grid Investigation Aim: My aim is to find out how the number of rows and columns in a square, on a certain sized grid will affect the difference between the product of the top left number and the bottom right number in the square subtracted from the top right number and the bottom left number. Prediction: I predict that when you've got numbers in a square box in a 10 by 10 grid, the difference will always be a square number because the box is shaped as a square. Method: Firstly, I'm going to work out the difference, in a 2 by 2 square in a 10 by 10 grid, between the two products that I get from multiplying the top left number and the bottom right number in the square by the top right number and the bottom left number. I am then going to repeat this another 3 times but these times I will work out a 3 by 3 square, 4 by 4 square and a 5 by 5 square. After doing this, I will try to find a formula that can find the difference in any sized square. I will then decrease the size of the grid so that it becomes 9 by 9 and will then do exactly the same method as I did before and then I will do this again but with a 5 by 5 grid. ...read more.

Middle

- (x� + 22x) (x� + 22x + 40) - (x� + 22x) x� + 22x + 40 - x� - 22x Difference =40 This proves algebraically that the difference in a 3 by 3 square is 40. I am now going to prove algebraically that in a 10 by 10 grid, with a 4 by 4 square the difference will be 90. x x+3 x+30 x+33 ((x + 3)(x + 30)) - x(x + 33) (x� + 3x + 30x + 90) - (x� + 33x) (x� + 33x + 90) - (x� + 33x) x� + 33x + 90 - x� - 33x Difference =90 This proves algebraically that the difference in a 4 by 4 square is 90. I am now going to prove algebraically that in a 10 by 10 grid, with a 5 by 5 square the difference will be 160. x x+4 x+40 x+44 ((x +4 )(x + 40)) - x(x + 44) (x� + 4x + 40x + 160) - (x� + 44x) (x� + 44x + 160) - (x� + 44x) x� + 44x + 160 - x� - 44x Difference =160 This proves algebraically that the difference in a 5 by 5 square is 160. I am now going to prove algebraically that in a 9 by 9 grid, with a 2 by 2 square the difference will be 9. ...read more.

Conclusion

x� + 18x + 45 - x� - 18x Difference =45 This proves algebraically that the difference in a 4 by 4 square is 45. I am now going to prove algebraically that in a 5 by 5 grid, with a 5 by 5 square the difference will be 80. x x+4 x+20 x+24 ((x + 4)(x + 20)) - x(x + 24) (x� + 4x + 20x + 80 ) - (x� + 24x) (x� + 24x + 80) - (x� + 24x) x� + 24x + 80 - x� - 24x Difference =80 This proves algebraically that the difference in a 5 by 5 square is 80. Now I'm furthering my investigation by looking at rectangles. This is a table to show the differences in squares in a 10 by 10 grid. No. of Rows (r) No. of Columns (c) Difference Formula 2 3 20 (10(r-1))x(c-1) 2 4 30 (10(r-1))x(c-1) 2 5 40 (10(r-1))x(c-1) 3 3 40 (10(r-1))x(c-1) 3 4 60 (10(r-1))x(c-1) 3 5 80 (10(r-1))x(c-1) 4 3 60 (10(r-1))x(c-1) 4 4 90 (10(r-1))x(c-1) 4 5 120 (10(r-1))x(c-1) 5 3 80 (10(r-1))x(c-1) 5 4 120 (10(r-1))x(c-1) 5 5 160 (10(r-1))x(c-1) In this formula, r is the No. of Rows and c is the No. of columns. Basically, you multiply (r-1) by 10 and then times your answer by (c-1). ?? ?? ?? ?? Sandeep Kutty 1 ...read more.

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