Higher Tier Task - Number Grid
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Use the following rule: find the product of the top left number and the bottom right number in the square. Do the same thing with the bottom left and the top right numbers in the square. Calculate the difference between these numbers.
INVESTIGATE!
I will begin with 2x2 windows on a 10x10 grid.
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34 x 25 = 850 49 x 58 = 2842 24 x 35 = 840- 48 x 59 = 2832-
10 10
88 x 97 = 8536 63 x 72 = 4536
87 x 98 = 8526- 73 x 62 = 4526-
10 10
3 x 12 = 36 6 x 15 = 90
2 x 13 = 26- 5 x 16 = 80-
10 10
For all of these windows you can see that the difference is always 10, so anywhere you put a 2x2 window on a 10x10 grid you will always get the same difference of 10. This can also be shown algebraically.
n
n+1
n+10
n+11
We can change this equation into: (n+1)(n+10)-n(n+11)
When multiplied outside of the brackets: n²+ 10n + n + 10 - n²-11n
This can be simplified to: n² + 11n + 10 - n² - 11n = 10
The n² and 11n cancel each other out therefore you are left with 10.
Now I will do the same, but this time, I will try 3x3 windows and see whether the difference will remain the same or have changed.
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53 x 71 = 3763 45 x 63 = 2835
51 x 73 = 3723- 43 x 65 = 2795-
40 40
9 x 27 = 243 39 x 57 = 2223
7 x 29 = 203- 37 x 59 = 2183-
40 40
80 x 98 = 7840 64 x 82 = 5248
78 x 100 = 7800- 62 x 84 = 5208-
40 40
Now I Know that the as you increase the window size on the grid, the difference also increases.
For the 3x3 widows you can see the difference is 40. Therefore if you put 3x3 windows anywhere on a 10x10 grid the difference will be equal to 40. We can again show the algebraic method of working out the difference.
n
n+1
n+2
n+10
n+11
n+12
n+20
n+21
n+22
We can change this into an equation:
(n+2)(n+20)-n(n+22)
When Multiplied outside of the brackets: n²+ 20n + 2n + 40 - n²- 22n = 40
This can be simplified to: n²+ 22n + 40 - n²- 22n = 40
Again the n² and 22n cancel each other out therefore you are left with 40.
Now I will substitute a 3x3 window at random into this equation.
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8² + (8 x 20) + (8 x 2) + 40 - 8² - (22 x 8) = 40
(64 + 160 + 16 + 40) - (64 + 176) = 40
280 - 240 = 40
Now I am going to try 4x4 windows on a 10x10 grid. I think that the difference will increase from previous examples, however I am not sure by how much.
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45 x 72 = 3240 44 x 71 = 3124
42 x 75 = 3150- 41 x 74 = 3034-
90 90
9 x 36 = 324 68 x 95 = 6460
6 x 39 = 234- 65 x 98 = 6370-
90 90
4 x 31 = 124 34 x 61 = 2074
1 x 34 = 34- 31 x 64 = 1984-
90 90
The difference for a 4x4 window is always going to be, no matter where you place it on the grid, 90. We can again prove that the difference will always be 90, by using algebra.
n
n+1
n+2
n+3
n+10
n+11
n+12
n+13
n+20
n+21
n+22
n+23
n+30
n+31
n+32
n+33
This can also be shown as an equation: (n+3)(n+30)-n (n+33)
When multiplied outside of the brackets: n²+ 30n + 3n + 90- n²- 33n = 90
This can be simplified to: n²+ 33n + 90- n²- 33n = 90
Once again n² and 33n both cancel each other out, therefore you are left with 90.
If I now substitute another random 4x4 window into the equation and I will hopefully get 90.
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2²+ (30 x 12) + (3 x 12) + 90 - 12² - (33 x 12)=90
44 + 360 + 36 + 90 - 144 - 396 = 90
630 - 540 = 90
I can now recognise a pattern here:
Square size Difference First Difference Second Difference
x1 0
10
2x2 10 20
30
3x3 40 20 50
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99
00
Use the following rule: find the product of the top left number and the bottom right number in the square. Do the same thing with the bottom left and the top right numbers in the square. Calculate the difference between these numbers.
INVESTIGATE!
I will begin with 2x2 windows on a 10x10 grid.
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
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00
34 x 25 = 850 49 x 58 = 2842 24 x 35 = 840- 48 x 59 = 2832-
10 10
88 x 97 = 8536 63 x 72 = 4536
87 x 98 = 8526- 73 x 62 = 4526-
10 10
3 x 12 = 36 6 x 15 = 90
2 x 13 = 26- 5 x 16 = 80-
10 10
For all of these windows you can see that the difference is always 10, so anywhere you put a 2x2 window on a 10x10 grid you will always get the same difference of 10. This can also be shown algebraically.
n
n+1
n+10
n+11
We can change this equation into: (n+1)(n+10)-n(n+11)
When multiplied outside of the brackets: n²+ 10n + n + 10 - n²-11n
This can be simplified to: n² + 11n + 10 - n² - 11n = 10
The n² and 11n cancel each other out therefore you are left with 10.
Now I will do the same, but this time, I will try 3x3 windows and see whether the difference will remain the same or have changed.
2
3
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9
0
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00
53 x 71 = 3763 45 x 63 = 2835
51 x 73 = 3723- 43 x 65 = 2795-
40 40
9 x 27 = 243 39 x 57 = 2223
7 x 29 = 203- 37 x 59 = 2183-
40 40
80 x 98 = 7840 64 x 82 = 5248
78 x 100 = 7800- 62 x 84 = 5208-
40 40
Now I Know that the as you increase the window size on the grid, the difference also increases.
For the 3x3 widows you can see the difference is 40. Therefore if you put 3x3 windows anywhere on a 10x10 grid the difference will be equal to 40. We can again show the algebraic method of working out the difference.
n
n+1
n+2
n+10
n+11
n+12
n+20
n+21
n+22
We can change this into an equation:
(n+2)(n+20)-n(n+22)
When Multiplied outside of the brackets: n²+ 20n + 2n + 40 - n²- 22n = 40
This can be simplified to: n²+ 22n + 40 - n²- 22n = 40
Again the n² and 22n cancel each other out therefore you are left with 40.
Now I will substitute a 3x3 window at random into this equation.
8
9
0
8
9
20
28
29
30
8² + (8 x 20) + (8 x 2) + 40 - 8² - (22 x 8) = 40
(64 + 160 + 16 + 40) - (64 + 176) = 40
280 - 240 = 40
Now I am going to try 4x4 windows on a 10x10 grid. I think that the difference will increase from previous examples, however I am not sure by how much.
2
3
4
5
6
7
8
9
0
1
2
3
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83
84
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98
99
00
45 x 72 = 3240 44 x 71 = 3124
42 x 75 = 3150- 41 x 74 = 3034-
90 90
9 x 36 = 324 68 x 95 = 6460
6 x 39 = 234- 65 x 98 = 6370-
90 90
4 x 31 = 124 34 x 61 = 2074
1 x 34 = 34- 31 x 64 = 1984-
90 90
The difference for a 4x4 window is always going to be, no matter where you place it on the grid, 90. We can again prove that the difference will always be 90, by using algebra.
n
n+1
n+2
n+3
n+10
n+11
n+12
n+13
n+20
n+21
n+22
n+23
n+30
n+31
n+32
n+33
This can also be shown as an equation: (n+3)(n+30)-n (n+33)
When multiplied outside of the brackets: n²+ 30n + 3n + 90- n²- 33n = 90
This can be simplified to: n²+ 33n + 90- n²- 33n = 90
Once again n² and 33n both cancel each other out, therefore you are left with 90.
If I now substitute another random 4x4 window into the equation and I will hopefully get 90.
2
3
4
5
22
23
24
25
32
33
34
35
42
43
44
45
2²+ (30 x 12) + (3 x 12) + 90 - 12² - (33 x 12)=90
44 + 360 + 36 + 90 - 144 - 396 = 90
630 - 540 = 90
I can now recognise a pattern here:
Square size Difference First Difference Second Difference
x1 0
10
2x2 10 20
30
3x3 40 20 50
