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  • Level: GCSE
  • Subject: Maths
  • Word count: 2272

Number grid

Extracts from this document...

Introduction

Mathematics Coursework; Number Grid

I have been asked to investigate a mathematical problem, in this piece of work I will explore the problem investigating a number of variables, using algebra to prove that the results hold for any combination of number.

12 x 23 = 276

13 x 23 = 286

The difference between the above two products is equal to 10.

I will repeat this using different numbers to see if there is a recurring pattern.

35 x 46 = 1610

36 x 45 = 1620

The difference between the above two products is equal to 10.

I predict that if I work out the following, the difference will be equal to 10.

38 x 49 =

39 x 48 =

I will now check to see if I am correct.

38 x 49 = 1862

39 x 48 = 1872

I was correct there is a difference of 10.

I am now going to try to prove that this holds for any 2 by 2 square using algebra.

Let n be the top left number and n + 11 be the bottom right number.

n(n+11) = n²+11n

Let n + 1 be the top right number and n + 10 be the bottom left number.

(n+1)(n+10) = n²+n+10n+10

              = n²+11n+10

I will now subtract the two algebraic products to get an overall product.

n²+11n+10

n²+11n       -

        +10

This proves that the difference is always 10



I have proved that the above results hold for any 2 by 2 square.

This shows that for any 2 by 2 square the difference will always be equal to 10.

To extend this investigation I am going to investigate the effect of changing the shape of the box.

...read more.

Middle

This shows that the difference is 60, I will use a 2 by 7 rectangle to see if I am correct.

1 x 17 = 17

7 x 11 = 77

The difference between the above two products is equal to 60, I was correct.

I will now use algebra to try and prove that this holds for any 2 by p rectangle.



Let n be the top left number and n + p + 9 be the bottom right number.

n(n+p)-9 = n²+np+9n

Let n + p - 1 be the top right number and n + 10 be the bottom left number.

(n+10)(n+p-1) = n²+np-9n+10P-10

I will now subtract the two algebraic products to get an overall product.

n²+np-9n+10P-10

n²+np+9n           -

            10p-10

This proves that the difference is always 10P-10


The above algebra tells me that for any 2 by p square the difference is always 10P-10.

To extend the investigation further I am going to investigate the effect of changing the size of the square, I will begin with a 3 by 3 square.

35 x 57 = 1995

37 x 55 = 2035

The difference between the above two products is equal to 40.

I am now going to try to prove that this holds for any 3 by 3 square using algebra.

Let n be the top left number and n + 22 be the bottom right number.

n(n+22) = n²+22n

Let n + 2 be the top right number and n + 20 be the bottom left number.

(n+2)(n+20) = n²+2n+20n+40

              = n²+22n+40

I will now subtract the two algebraic products to get an overall product.

n²+22n+40

n²+22n       -

        +40

...read more.

Conclusion

(n+1)(n+w) = n²+nw+w+1

I will now subtract the two algebraic products to get an overall product.

n²+nw+w+1

n²+nw+1    -

          +w

This proves that the difference is always equal to w

I have proved that the above results hold for any n by w square. W is equal to the size of the overall number grid.

It shows that the rule applies to any combination of letters or numbers, within its limits, there are certain limitations which mean that the solution will not hold for every number, it will not hold for any grid which contains 100 or more numbers, (in other words greater than 10 by 10), this shows that although that the solution does hold for most possibilities certain limitations mean that it cannot fit for absolutely every possibility.

I will now co-ordinate all three of my variables to see if there is any link, when I changed the size of square I discovered that the difference of any m by m square was equal to 10 (p-1)², when I changed 10 (p-1), already I can spot that the only difference between these two differences is that the square difference is squared, and the rectangle difference isn’t.

In conclusion, I have learnt that the by simply exploring and investigating different variables you can discover that there are certain links and connected outcomes, from my investigations I have discovered that the differences that I have found, hold for that particular shape or grid size, I have proved this algebraically, providing the general solution which shows that it holds for any number or letter.

Robert Shaw

Candidate Number: 1075 Centre Number: 41284

...read more.

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