GCSE Maths coursework

Cross Numbers

Here is a number square.

It shows all the numbers from 1 to 100.        

I can form cross numbers by placing a cross on this grid.

I will use the cross shown below.

In this investigation, I will try to find a master formula for various functions that I use with a certain cross shape.

To do this, I will follow the procedure given below:

1. I will then pick a number on the grid, (labelled as X).

2. Use a mathematical formula using the 4 numbers around it.                           E.g. (24+33)-(22+13).

3. When an answer is found, I will repeat the formula but pick a different value for X. Repeat this 3 times. If the same answer is achieved, then I will assign an algebraic formula for each number around X.

4. Once I have assigned an algebraic formula for each square, I will replace the numbers in the formula with the algebraic form and justify this mathematically.

5. I will also be changing the horizontal grid value (g) to further my investigation.

6. This will therefore give me a master formula.

I will use the following symbols

X=center number.

g=number of squares across the horizontal axis.

e.g.

g=10

Prediction

I predict that whichever grid size I use the number above always = X-g, the number below always = X+g, the number on the left always = (X-1)

X+g

[(X-1) + (X+1)] + [(X-g) + (X+g)]

        = X-1 + X+1 + X-g + X+g

        = 4x

If I replace the centre number, I get the following results.

If X=87 then

[(87-1) + (87+1)] + [(87-4) + (87+4)]

        = 87-1 + 87+1 + 87-4 + 87+4

        = 4x87

        = 348

If X=52 then

[(52-1) + (52+1)] + [(52-4) + (52+4)]

        = 52-1 + 52+1 + 52-4 + 52+4

        = 4x52

        = 208

If X=16 then

[(16-1) + (16+1)] + [(16-4) + (16+4)]

        = 16-1 + 16+1 + 16-4 + 16+4

        = 4x16

        = 64

This tells me that whenever I use the above formula, the solution is always 4X ,and if I replace X with any number (apart from the outside edge numbers) I always get 4xX as an answer.

 Therefore this is a master formula for this shape, and grid sizes 10x10, 6x10 and 4x10.

c)

        [(X+g) - (X-g)] – [(X+1) - (X-1)]

        = [X+g – X+g] – [X+1 – X+1]

        = 2g – 2

If I replace X, I get the following results.

If X = 33 then

[(33+4) - (33-4)] – [(33+1) - (33-1)]

        = [33+4 – 33+4] – [33+1 – 33+1]

        = 2x4–2

        =6

If X=83 then

[(83+4) - (83-4)] – [(83+1) - (83-1)]

        = [83+4 – 83+4] – [83+1 – 83+1]

        = 2x4–2

        = 6

If X=49

[(49+4) - (49-4)] – [(49+1) - (49-1)]

        = [49+4 – 49+4] – [49+1 – 49+1]

        = 2x4–2

        = 6

This tells me that whenever I use the above formula, the solution is always 2g – 2, hence I always get 4 as an answer. Therefore this is a master formula for this shape and this grid.

Overall so far, the only formula that works for all three grid sizes (4x10, 6x10 and 10x10) to give the same answer is:

[(X-1) + (X+1)] + [(X-g) + (X+g)]

= X-1 + X+1 + X-g + X+g = 4x (addition formula)

All other formulas work but give a different figure but the same algebraic solution.

For the subtraction formula the solution is g²-1 and for multiplication it is 2g-2.

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Proving that all Algebraic formulas work with all grid sizes

Grid size 4x10:

If  X=10 and g=4

The number above x is X-g because 10-4=6  

The number below x is X+g because 10+4=14 .

The number to the left is (X-1) because

14-1=13.

The number to the right is (X+1) because                                         14+1=15.

Grid size 8x10

If  X=10 and g=4

The number above x is X-g because 20-8=12 and g=8.

The number below x is X+g: 20+8=28 and g=8

The number to the  left is (X-1) because 20-1=19.

Grid size 6x10

       If  X=39 and g=6

                     The number to the right is (X+1)  because 20+1=21

The number above x is X-g because 39-6=33 and g=6.

The number below x is X+g because 39+6=45 and g=6.

The number to the left is (X-1) because 39-1=38.

The number to the right is (X+1)     because 39+1=40.

This shows that these

 are the master formulas for all values of (g)  with this shape.

To show that this works, If X=8 for g=4 then:

Grid size 8x10

If X=14

The top left will always be X-(g+1) because 14-(8+1) = 5

The top right will always be X-(g-1) because 14-(8-1)= 7

The bottom left will always equal X+(g-1) because 14+(8-1)=21

The bottom right will equal X+(g+1) because 14+(8+1)=23

Grid size 4x10

If X=14

The top left will always be X-(g+1) because 14-(4+1) = 9

The top right will always be X-(g-1) because 14-(4-1)= 11

The bottom left will always equal X+(g-1) because 14+(4-1)=17

The bottom right will equal X+(g+1) because 14+(4+1)=19

 

This shows that these

 are the master formulas for all grid size variations with this shape.

Conclusions:

In this investigation I found several different master formulas and one Universal formula. These are summarised in the table below.

g= grid size , X= center number of a cross

A master formula is a formula that works for a specific shape on all the three grid sizes ( 10 x 10, 6 x10 and 4 x 10) that I investigated.

A universal formula is a formula that can be used for all shapes and all grid sizes that I investigated.

I would have liked to further my investigation by using a three dimensional grid.