# Number Grid Investigation

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Introduction

Page No.1 Number Grid Investigation I have been asked to investigate on a number grid 10 wide 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 2x2 grids 6 7 16 17 1. a) My first 2x2 grid is (6 x 17) - (7 x 16) = -10 18 19 28 29 b)My second 2x2 grid is (18 x 29) - (19 x 28) = -10 2 3 12 13 c)My third 2x2 grid is (2 x 13) - (3 x 12) = -10 24 25 34 35 d)My fourth 2x2 grid is (24 x 35) - (25 x 34) = -10 2. I predict that in the next grid the answer will be -10. 15 16 25 26 3 Test prediction- I will use the grid: (15 x 26) - (16 x 25) = -10 4. Proving the rule Y Y+1 Y + 10 Y + 11 (Y x Y + 11) - (Y + 1 x Y + 10) = (Y2 + 11) - (Y2 + 11) = 0 ` Page No.2 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 3x3 grids 14 15 16 24 25 26 34 35 36 1.a) ...read more.

Middle

Predicting an N*N Grid I predict that the answer to any nxn grid will be -10 x (n-1) � as long as 2<n<10. I.E. X X+ (n-1) X + 10(n-1) X + 11(n-1) Here, X=2 and N=3. This gives us... X[X + 11(n-1)] - [(X + n-1)(X+10(n-1)] Removing the [ ] s... X(X+11N-11) - (X+N-1) (X+10N-10) Which looks like this with the ( ) s removed... X�+11NX-11x - (X�+10NX-10X+NX-X+10N�-10N-10N+10) Then... -10N�+20N-10 And... -10(N�-2N+1) Which equals... -10(n-1)� Rectangles Page No. 7 1. 2 x 3 2 x 4 2 x 5 1 2 3 11 12 13 1 4 11 14 1 5 11 15 (1 x 13) - (13 - 33) = (1 x 14) - (4 x 11) (1 x 15) - (5 x 11) 13- 33 = =14 - 44 =15 - 55 -20 = -30 = -40 1 6 21 26 3 x 4 3 x 5 3 x 6 1 4 21 24 1 5 21 25 (1 x 24) - (4 x 21) (1 x 25) - (5 x 21) (1 x 26) - (6 x 21) = 24 - 84 =25 - 105 =26 - 126 =-60 = -80 = -100 1 7 31 37 4 x 5 4 x 6 4 x 7 1 5 31 35 1 6 31 36 (1 x 35) - (5 x 31) (1 x 36) - (6 x 31) (1 x 37) - (7 x 31) = 35 - 155 = 36 - 186 = 37 - 217 = -120 = -150 = - 180 2 x 3 = -20 2 x 4 = -30 2 x 5 = -40 3 x 4 = -60 3 x 5 = -80 ...read more.

Conclusion

+a (n-1)]} - {[x+ (n-1)] [x+a (n-1)]} Taking the out '{ }'s leave... = [x(x+n-1+an-a)] - [(x+n-1) (x+an-a)] Removing the '[ ]'leaves... = (x�+nx-x+anx-ax) - (x�+anx-ax+nx+anm-an-x-an+a) Extracting the last brackets leaves... = - (anm-an-an+a) And again simplifies to... =-anm+an+an-a =-a (nm-2n+1) This finishes as... -a (n-1�) This proves my hypothesis. Finding a formula for an M*N grid on an A*A Master grid. As explained in the end of the rectangles section we can include an 'm' by removing an 'n'. In this formula 'm' equals width, 'n' equals the amount of rows down, 'a' equals the size of the Master grid and x equals the starting value on the grid. x X+(m-1) X+a(n-1) X+a(n-1)+(m-1) = {x[x+ (m-1) +a (n-1)]} - {[x+ (m-1)] [x + a (n-1)]} = [x(x+m-1+an-a)] - [(x+m-1) (x + an-a)] = (x�-x+mx+anx-ax) - (x�+anx-ax+mx+anm-am-x-an+a) = - (anm-am-an+a) = - anm+am+an-a = -a (nm-m-n+1) = -a (n-1) (m-1) This was achieve by replacing the 10, which corresponds to the size of the Master grid, with an 'a' using the working form rectangle. If we replace the 10 with the 'a's it doesn't affect the formula it only means that everywhere a 10 appears we see an 'a' instead. Page No. 13 This formula is identical to the rectangular one on a 10*10 Master grid but with an 'a' instead of an -10. This is because it is the same but we have an 'a' representing the size of the master grid which in the 10*10 master grid was 10. But because on squares they have an equal length/width we don't need an 'm' to represent the width. Leaving the formula as -a (n-1�) which means - (size of the Master grid) * (length-1�) which is similar to -10(n-1�). ?? ?? ?? ?? Sam Fiske ...read more.

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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