6(2 – 2)²
6(4²) = 24
Now my formula was complete and correct. Finally I moved onto cubes which had no faces painted.
0 1 8 27 64
1 7 19 37
6 12 18
6 6
This time I found that the third difference was constant. This meant the formula contained ‘n³’. I noticed the original numbers were all cube numbers, so I tried the formula n³ by using ‘n = 2’.
n = 2
2 x 2 x 2 = 8
This was wrong. The answer I was looking for was 0. As I had used it in my previous formulas (n – 2), I decided to see if it would work again.
n = 2
(2 – 2) x (2 – 2) x (2 – 2) = 0
This formula worked which meant I had now obtained all my formulas. Just to check them, I tested them on a cube sized 10 x 10 x 10. I drew the diagram, and started counted the cubes as I was now confident enough to use the drawing rather than building the cube. Then I predicted the number of cubes with the faces.
For a cube sized 10 x 10 x 10…
The number of cubes with three faces painted will be 8.
The number of cubes with two faces painted will be 96.
12(n – 2)
12(10 – 2)
12 x 8 = 96
The number of cubes with one face painted will be 384.
6(n – 2)²
6(10 – 2)²
6 x 64 = 384
The number of cubes with zero faces painted will be 512.
(n – 2)³
(10 – 2)³
8 x 8 x 8 = 512
After I had predicted the numbers, I counted the faces on the cube, and found that all of them were correct.
As I had now found the formulas, I could predict the numbers of faces for the original question…
For a cube sized 20 x 20 x 20…
The number of cubes with three faces painted will be 8.
The number of cubes with two faces painted will be 216.
12(n – 2)
12(20 – 2)
12 x 18 = 216
The number of cubes one face painted will be 1944.
6(n – 2)²
6(20 – 2)²
6(18²)
6 x 324 = 1944
The number or cubes with zero faces painted will be 5832.
(n – 2)³
(20 – 2)³
18³ = 5932
As I looked over my results table again, I noticed that all the numbers of faces for each, added up to the total number of cubes altogether.
For example:
3 x 3 x 3 = 27 (8 + 12 + 6 + 1)
Three faces – 8
Two faces – 12
One face – 6
Zero Faces – 1
Therefore, if adding all the faces together, equalled the total of the number cubed, I predicted that by adding all the formulas together, would also cancel down to be n³. This would be because there are three dimensions to a cube. I investigated further into this.
Adding all the formulas = 8 + 12(n – 2) + 6(n – 2)² + (n – 2)³
12(n – 2) = 12n – 24
6(n – 2)² = 6(n – 2)(n – 2)
= 6n² – 12n – 12n + 24
= 6n² – 24n + 24
(n – 2)³ = (n – 2)(n – 2)(n – 2)
= (n – 2)(n² – 2n – 2n + 4)
= n³ – 2n² – 2n² + 4n – 2n² + 4n + 4n – 8
= n³ – 6n² + 12n – 8
Totalling up all the formulas (after expanding out the brackets) gave me my final answer…
(8) + (12n - 24) + (6n² - 24n + 24) + (n³ - 6n² + 12n - 8)
Cancelling out numbers within the formula left me with n³ as I predicted.
Reasons why my formulas worked
A reason why the term ‘n – 2’ was always used in the formulas was as I looked at the cube sizes, I noticed that the terms (n) started at 2 and not at 0. They started at 2 x 2 x 2 and not 0 x 0 x 0, and so I subtracted 2.
For cubes with two faces, I used the number 12 because the numbers went up in multiples of twelve.
For cubes with three faces, I used the number 6 because the formula contained ‘n²’ and so I had to divide the second difference (12) by 2.
For cubes with 0 faces painted, I used the term ‘n³’ because the numbers were the cube numbers.
EXTENSION 1
For my first extension, I investigated further into the number of faces painted, but this time in cuboids.
For these cuboids I am going to keep two of the lengths constant, and change one (2 x 2 x 2, 2 x 2 x 3, 2 x 2 x 4, 2 x 2 x 5, 2 x 2 x 6, 2 x 2 x 7, 2 x 2 x 8, 2 x 2 x 9).
Again I drew diagrams, and began counting the faces of the cubes…
Results Table
Patterns and Formulas
As I looked at my results, I realised I only had to find formulas for cubes with three face and two faces painted, as the others had a constant zero.
I saw that again there was no formula for three faces painted as it was the same constant 8.
Now, I looked for patterns within the results of cubes with two faces painted.
0 4 8 12 16
4 4 4 4
I found the first difference was a constant 4, and so I thought the formula may be ‘4n’. I again tried it out on 2 x 2 x 3…
4n
4 x 3 = 12
This was wrong, as it was too big. The answer I needed was four, and so I tried subtracting 8. This was right, but just in-case I tested it on 2 x 2 x 4.
(4 x 4) – 8
16 – 8 = 8
This was correct. It gave me the right answer, but I still needed to test the formulas for a final time. This time, I drew a cube sized ‘2 x 2 x 10’…
For a cube sized 2 x 2 x 10…
The number of cubes with three faces painted will be 8.
The number of cubes with two faces painted will be 32.
4n – 8
(4 x 10) – 8
40 – 8 = 32
These answers were correct, which meant that my formulas were as well.
EXTENSION 2
For my second extension, I am going to investigate cuboids again but this time a different size to the previous ones.
Again, I shall firstly draw the cubes, and then carry on as per usual…
Results Table
Patterns and Formulas
This time, I had to find formulas for three faces painted, two faces painted, and one face painted only.
Again, there was no pattern for three faces painted, except a constant 8. I realised that this number wasn’t going to change, so I started looking for a pattern between the numbers for two faces painted.
4 8 12 16 20
4 4 4 4
This time, the difference was the same as before. I thought about trying ‘4n’ but as I calculated it in my head, it was not going to work. I realised I would have to minus 4 to get my answer. I tested this out using 2 x 3 x 3.
(4 x 3) – 4
12 – 4 = 8
This formula was correct. I now had to find one for two faces painted…
0 2 4 6 8
2 2 2 2
Now, the first difference was two, and so I decided to take the obvious approach, and think the formula was ‘2n’. I tested this on 2 x 3 x 5…
2n
2 x 5 = 10
This was incomplete as the answer I needed was 6. I chose to subtract 4 from there, and see if the formula worked now.
2 x 5 = 10
10 – 4 = 6
Now I had obtained all the formulas, I again tested them on 2 x 3 x 10. I drew the cube, and started predicting the different number of faces.
For a cube sized 2 x 3 x 10…
The number of cubes with three faces painted will be 8.
The number of cubes with two faces painted will be 36.
4n – 4
(4 x 10) – 4
40 – 4 = 36
The number of cubes with one face painted will be 16.
2n – 4
(2 x 10) – 4
20 – 4 = 16
After my prediction, I counted the cubes, and they were all correct. I had all my formulas and was ready to move on one step further.
EXTENSION 3
When I investigated cuboids previously, I kept two of my lengths constant. This did not give many formulas, and therefore I am going to try keeping all my lengths different from each other. This would help me explore cubes and cuboids of different sizes.
From here, I hope to achieve a formula that will enable me to find the number of faces on a cube/cuboid any size.
Results Table
Patterns and Formulas
This time, as I was trying to find a general formula that would work for any sized cube or cuboids; I was not going to find specific formulas. Instead, I was going to look at the shape of the cube to see if it had any connection to the number of faces. I labelled the three lengths were labelled ‘a’, ‘b’ and ‘c’.
No matter what size cube/cuboid it was, the corners (three faces painted) would always be eight (8).
Then I noticed that for two faces painted, it would be a side (for example ‘a’) minus two cubes for the corners (three faces). The remaining cubes on the edges would all be two faces. Then the formula (a – 2) would have to be multiplied by four to become 4(a – 2) as it is a 3-d cube (so all the parallel edges would total up to four). The same would then apply to sides ‘b’ and ‘c’. They would become 4(b – 2) and 4(c – 2).
Together, the formula for 2 faces would be…
4(a – 2) + 4(b – 2) + 4(c – 2)
Now, I had to find a formula for cubes with one face. I noticed that these particular cubes were on the faces of the cube. As I looked for the size of the cubes, I realised that it was ‘a – 2’. I found that I needed to find another length to be able to get the area of cubes with one face. I observed the area would be (a – 2)(b – 2). As there was two of the same faces (the opposite face), I would have to multiply my formula by two, making it 2(a – 2)(b – 2). This was the first part of my formula, and I had another two to find. It would be easier as I knew how to find them. The second part of the formula would be 2(a – 2)(c – 2) and lastly it would be 2(b – 2)(c – 2).
Therefore, the total formula for one face would be:
2(a – 2)(b – 2) + 2(a – 2)(c – 2) + 2(b – 2)(c – 2)
Now to complete the investigation, I had to find a formula for cubes with zero faces. This was a problem as cubes with zero faces were hidden away inside and therefore they could not be seen. I had to imagine that all the outer surfaces had been chopped off, (including one layer form the top and bottom). This would count as my three faces, two faces and one face. I was soon to realise that the length was a side minus two (a – 2, b – 2 and c – 2). Hence if I multiplied the three together, I would get the total.
This meant my formula for zero faces would be:
(a – 2)(b – 2)(c – 2)
Cube showing the dimensions and sizes...
Just to make sure that these formulas were correct, I decided to check them.
For a cuboid sized 7 x 8 x 9…
The number of cubes with three cubes painted will be 8.
The number of cubes with two cubes painted will be 72.
4(a – 2) + 4(b – 2) + 4(c – 2)
4(7 – 2) + 4(8 – 2) + 4(9 – 2)
(4 x 5) + (4 x 6) + (4 x 7)
20 + 24 + 28 = 72
The number of cubes with one face painted will be 214.
2(a – 2)(b – 2) + 2(a – 2)(c – 2) + 2(b – 2)(c – 2)
2(7 – 2)(8 – 2) + 2(7 – 2)(9 – 2) + 2(8 – 2)(9 – 2)
2(5 x 6) + 2(5 x 7) + 2(6 x 7)
(2 x 30) + (2 x 35) + (2 x 42)
60 + 70 + 84 = 214
The number of cubes with zero faces painted will be 210.
(a – 2)(b – 2)(c – 2)
(7 – 2)(8 – 2)(9 – 2)
5 x 6 x 7 = 210
Now to see whether my formulas would work on any sized cubes/cuboids, I decided to test them on a cuboid sized 3 x 5 x 9…
For a cube sized 3 x 5 x 9…
The number of cubes with three faces painted will be 8.
The number of cubes with two faces painted will be 44.
4(3 – 2) + 4(5 – 2) + 4(9 – 2)
(4 x 1) + (4 x 3) + (4 x 7)
4 + 12 + 28 = 44
The number of cubes with one face painted will be 62.
2(3 – 2)(5 – 2) + 2(3 – 2)(9 – 2) + 2(5 – 2)(9 – 2)
2(1 x 3) + 2(1 x 7) + 2(3 x 7)
(2 x 3) + (2 x 7) + (2 x 21)
6 + 14 + 42 = 62
The number of cubes with zero faces painted will be 21.
(3 – 2)(5 – 2)(9 – 2)
1 x 3 x 7 = 21
This was correct which also meant my formulas were as well.
As done previously, I wanted to see whether or not the formulas totalled up, equalled ‘abc’ (the three dimensions of the cuboid). If adding up all the formulas came to n³, then calculating up all the general formulas should equal ‘abc’. Like before, I took all the formulas I gained, and added them together, from which I could cancel the numbers down.
4(a – 2) + 4(b – 2) + 4(c – 2)
= 4a – 8 + 4b – 8 + 4c – 8
= 4a + 4b + 4c – 24
2(a – 2)(b – 2) + 2(a – 2)(c – 2) + 2(b – 2)(c – 2)
= (2ab – 4a – 4b + 8) + (2ac – 4a – 4c + 8) + (2bc – 4b – 4c + 8)
= 2ab + 2ac + 2bc – 8a – 8b – 8c + 24
(a – 2)(b – 2)(c – 2)
= (c – 2)(ab – 2a – 2b + 4)
= abc – 2ac – 2bc + 4c – 2ab + 4a + 4b – 8
= abc – 2ab – 2ac – 2bc + 4a + 4b + 4c – 8
The total for adding each of the formulas is…
[8] + [4a + 4b + 4c – 24] + [2ab + 2ac + 2bc – 8a – 8b – 8c + 24] + [abc – 2ab – 2ac – 2bc + 4a + 4b + 4c – 8]
After cancelling out all my numbers, I found again, that my formulas all came down to ‘abc’. This proved that my prediction was right again, and that the basis of all the formulas was the three different measurements I started off with… ‘a’, ‘b’ and ‘c’.
After finding this out, I proposed the question of whether:
A cube is a special form of a cuboid?
To do this, I would have to take values of a cube (all three lengths constant), and substitute them into my general formulas that I found for my cuboids. Firstly, I would test the cube with sides 3 x 3 x 3…
Three faces - 8
Two faces - 4(a – 2) + 4(b – 2) + 4(c – 2)
4a + 4b + 4c – 24
(4 x 3) + (4 x 3) + (4 x 3) – 24
12 + 12 + 12 – 24 = 12
One Face - 2(a – 2)(b – 2) + 2(a – 2)(c – 2) + 2(b – 2)(c – 2)
2ab + 2ac + 2bc – 8a – 8b – 8c + 24
(2 x 3 x 3) + (2 x 3 x 3) + (2 x 3 x 3) – (8 x 3) – (8 x 3) – (8 x 3) + 24
18 + 18 + 18 – 24 – 24 – 24 + 24 = 6
Zero Faces - (a – 2)(b – 2)(c – 2)
abc – 2ab – 2ac – 2bc + 4a + 4b + 4c – 8
(3³) – (2 x 3 x 3) – (2 x 3 x 3) – (2 x 3 x 3) + (4 x 3) + (4 x 3) + (4 x 3) – 8
27 – 18 – 18 – 18 + 12 + 12 + 12 – 8 = 1
The values I got for this cube were all correct. Just to be sure, I checked once again, with the cube sized 7 x 7 x 7
Three faces - 8
Two faces - 4(a – 2) + 4(b – 2) + 4(c – 2)
4a + 4b + 4c – 24
28 + 28 + 28 – 24 = 60
One face - 2(a – 2)(b – 2) + 2(a – 2)(c – 2) + 2(b – 2)(c – 2)
2ab + 2ac + 2bc – 8a – 8b – 8c + 24
98 + 98 + 98 – 56 – 56 – 56 + 24 = 150
Zero faces - (a – 2)(b – 2)(c – 2)
abc – 2ab – 2ac – 2bc + 4a + 4b + 4c – 8
343 – 98 – 98 – 98 + 28 + 28 + 28 – 8 = 125
This was again correct, which meant that my theory was also right. A cube was a special form of a cuboid, and this was demonstrated by my general formulas working for my cubes.
Conclusion
Consequently, I did find a general formula that would work for any sized cube or cuboid. In the process, I also found formulas for keeping all the lengths constant, two same numbers as constants, and two different numbers for constants…
n x n x n (All lengths constant)
Three faces – 8
Two faces – 12(n – 2)
One face – 6(n – 2)²
Zero faces – (n – 2)³
TOTAL FORMULA – 8 + 12(n – 2) + 6(n – 2)² + (n – 2)³
2 x 2 x n (Two same numbers constant)
Three faces – 8
Two faces – 4n – 8
One face – 0
Zero faces – 0
TOTAL FORMULA – 8 + (4n – 8)
2 x 3 x n (Two different numbers constant)
Three faces – 8
Two faces – 4n – 4
One face – 2n – 4
Zero faces – 0
TOTAL FORMULA – 8 + (4n – 4) + (2n – 4)
a x b x c (All lengths different)
Three faces – 8
Two faces – 4(a – 2) + 4(b – 2) + 4(c – 2)
One face – 2(a – 2)(b – 2) + 2(a – 2)(c – 2) + 2(b – 2)(c – 2)
Zero faces – (a – 2)(b – 2)(c – 2)
= + + + +
