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Maths Opposite corners

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Introduction

Maths Coursework

Opposite Corners

        My task is to investigate a number grid. If you take a 2x3 rectangle and place it on a 10x10 number grid the diagonal difference of the numbers inside is 20. I want to first investigate whether the diagonal difference is always 20, no matter where the rectangle is, and prove this. I will then investigate further by changing the size of the number grid and of the rectangle.

image00.png First I investigated two rectangles to see if the diagonal difference of both was 20.

image01.png

image11.png Both of these diagonal differences were 20.

        I decided to make the rectangle larger by one square wider and one deeper to see what happens.

image12.pngAgain I investigated two rectangles to see if the diagonal difference is the same for both.

image13.png

Both came out as 60 so the formula is:

x²+23x-(x²+60+23x)=60

So far the diagonal differences are both factors of twenty, meaning the first digit is an even number (When there are two digits) and the number ends in zero.

...read more.

Middle

 Both these were 60. So the diagonal difference has risen by 20 again. The diagonal differences are obviously going up in a sequence.

        I decided to make a formula for the diagonal difference of my first box by calling the number in the top left corner ‘x’:

image03.png

        If I call the number in the top left hand corner ‘x’ then I can work out the rest of the numbers. The top right hand number has to be ‘x+2’. This is because in a 2X3 box this number will always be 2 more than the one on the top left. The bottom left will be ‘x+10’ because every step down in a 10X10 grid adds 10 to the number. The bottom right must be ‘x+12’ because it is two higher than ‘x+10’. I multiply ‘x’ with ‘x+12’ just like before and do the same with ‘x+2’ and ‘x+10’. The products of these numbers are:

        x(x+12)=x²+12x

        (x+2)(x+10)=x²+20+12x

The difference between these numbers is:

        x²+12x-(x²+20+12x)=20

This shows that the answer is always 20.

...read more.

Conclusion

(x+m-1)(x+pn-p)-x(x+pn+m-p-1)= x²+xpn-xp+mpn+mx-mp-x-pn+p-x²-xpn-xm+xp+x

                                           = pnm-pn-pm+p

My final formula for the diagonal difference of a rectangle/square of any size on a grid of any size is pnm-pn-pm+p. This proves my prediction that ‘p’ will replace the number ten in my earlier formula. I’ll prove this by using the formula on my first rectangle:

image08.png m=3, n=2, p=10.    60-20-30+10=20    The diagonal difference is 20.

        I will now try my formula with a 4X5 rectangle on a 20X20 grid to prove that it works with different sized grids and boxes.

image10.png

Diagonal difference= 1X84=84

= 4X81=324

                        =324-84

        = 240

Diagonal difference= pnm-pn-pm+p

                        = 20X5X4-20X5-20X4+20

                        = 240

...read more.

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