Son's Maths doing our heads in... HELP
20 October 2004, 10:19 PM
#1
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Son's Maths doing our heads in... HELP
Here's what we've got to fathom....
Need to work out a formula for the total amount of cubes in a 3-D shape....
Shape is like this...
The cubes on the outside of shape (or crust if you like) are coloured in black but the ones on the insed that can't be seen are white. the formula must be some way of describing the total number of cubes in relation to the black and white cubes and if necessary the dimensions of the shape.
HELP PLEASE
Need to work out a formula for the total amount of cubes in a 3-D shape....
Shape is like this...
The cubes on the outside of shape (or crust if you like) are coloured in black but the ones on the insed that can't be seen are white. the formula must be some way of describing the total number of cubes in relation to the black and white cubes and if necessary the dimensions of the shape.
HELP PLEASE
20 October 2004, 11:07 PM
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Son worked out the 2-D part of the question OK (I hope!)
t=Total
d= Dimensions
formula is t=d^2 + (d-1)^2
for example for a shape that is a 4 x 4 (d x d) cross shape of cubes the formula is
t=4^2 + (4-1)^2
t=16 + (3)^2
t=16 + 9
t= 25
3-d version is way beyond my little grey cells
t=Total
d= Dimensions
formula is t=d^2 + (d-1)^2
for example for a shape that is a 4 x 4 (d x d) cross shape of cubes the formula is
t=4^2 + (4-1)^2
t=16 + (3)^2
t=16 + 9
t= 25
3-d version is way beyond my little grey cells
20 October 2004, 11:13 PM
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I've had a few cans watching the foota, but here's my best shot:
I was never any good at writing thoeries, but someone may be able to put my babblings into a theory.
For me, you've got to think of each layer as a 2d shape and you'll have 2 separate square numbers added together for each layer. You then add them all together to get the total.
For example, a 3x3 will actually look like this:
# # #
.# #
# # #
.# #
# # #
(ignore the dots, they're just to bring things in line.)
Therefore, (the first shape doesn't really count), the second shape is:
top: 1 squared (1)
middle: 2 squared + 1 squared (4+1)
bottom: 1 squared (1)
Add them all together and you get 7 cubes.
The 3x3 is:
top: 1 squared (1)
2nd row: 2 squared + 1 squared (4+1)
middle row: 3 squared + 2 squared (9+4)
4th row: 2 squared + 1 squared (4+1)
bottom: 1 squared (1)
Total is 25 cubes.
4x4 is:
top: 1 squared (1)
2nd row: 2 squared + 1 squared (4+1)
3rd row: 3 squared + 2 squared (9+4)
4th row: 4 squared + 3 squared (16+9)
5th row: 3 squared + 2 squared (9+4)
6th row: 2 squared + 1 squared (4+1)
7th row: 1 squared (1)
Total cubes is 63.
For 5x5 you just add (5 squared to 4 squared) onto the 63 you got for 4x4.
Hope this helps and I hope someone can work out a formula for you.
Andy.
Edited to say:
Took so long for me to write it, someone beat me to it, and in a more concise way, lol
I was never any good at writing thoeries, but someone may be able to put my babblings into a theory.
For me, you've got to think of each layer as a 2d shape and you'll have 2 separate square numbers added together for each layer. You then add them all together to get the total.
For example, a 3x3 will actually look like this:
# # #
.# #
# # #
.# #
# # #
(ignore the dots, they're just to bring things in line.)
Therefore, (the first shape doesn't really count), the second shape is:
top: 1 squared (1)
middle: 2 squared + 1 squared (4+1)
bottom: 1 squared (1)
Add them all together and you get 7 cubes.
The 3x3 is:
top: 1 squared (1)
2nd row: 2 squared + 1 squared (4+1)
middle row: 3 squared + 2 squared (9+4)
4th row: 2 squared + 1 squared (4+1)
bottom: 1 squared (1)
Total is 25 cubes.
4x4 is:
top: 1 squared (1)
2nd row: 2 squared + 1 squared (4+1)
3rd row: 3 squared + 2 squared (9+4)
4th row: 4 squared + 3 squared (16+9)
5th row: 3 squared + 2 squared (9+4)
6th row: 2 squared + 1 squared (4+1)
7th row: 1 squared (1)
Total cubes is 63.
For 5x5 you just add (5 squared to 4 squared) onto the 63 you got for 4x4.
Hope this helps and I hope someone can work out a formula for you.
Andy.
Edited to say:
Took so long for me to write it, someone beat me to it, and in a more concise way, lol
Last edited by Walwal; 21 October 2004 at 12:06 AM.
20 October 2004, 11:29 PM
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Hi Guys thanks for helping here... is this really GCSE LOL
Walwal
should the 3rd row be : 3 squared + 2 squared (9+4) plus 1
same for the 4th row etc?
Loost
what does 'e' refer to? and does the 3 mean cubed...
I feel such a durrrrr brain!
Walwal
should the 3rd row be : 3 squared + 2 squared (9+4) plus 1
same for the 4th row etc?
Loost
what does 'e' refer to? and does the 3 mean cubed...
I feel such a durrrrr brain!
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20 October 2004, 11:38 PM
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Originally Posted by missyc
Hi Guys thanks for helping here... is this really GCSE LOL
Walwal
should the 3rd row be : 3 squared + 2 squared (9+4) plus 1
same for the 4th row etc?
Walwal
should the 3rd row be : 3 squared + 2 squared (9+4) plus 1
same for the 4th row etc?
A 2x2 has 4 cubes round the outside, a 1 in the middle. (i.e. 2x2 + 1x1)
A 3x3 has 8 round the outside, 1 in the middle (giving 3x3), then 4 filling in the gaps in between (giving the 2x2), (3x3 + 2x2)
And so on.
Hard to explain when I can't draw a picture I'm afraid.
Andy.
20 October 2004, 11:54 PM
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I havn't got time to look at this properly I'm afraid as I'm off to bed 
, but from a quick observation, the cubes are cumulative, i.e. the 1x1x1 fits inside and comprises the white cubes within the 2x2x2. Likewise the 2x2x2 shape makes up the white cubes in the 3x3x3 so the number of whites lags behind by one shape e.g.:
shape, black, white, total
1x1x1, 1, 0, 1
2x2x2, 6, 1, 7
3x3x3,18, 6, 24
4x4x4, 38, 18, 56
You have the formula for the total in one layer (the widest layer in the shape that defines its dimensions), in order to get the whole shape you need to work it out thus:-
where Tn = total blocks in a layer of dimensions n x n
Total for shape = Tn + 2(Tn-1) + 2(Tn-2).... +2(T1) (to make up the shape there is the main layer in the middle and then 2 layers of each dimension, one on each side of the main layer for each dimension all the way down to one (every shape has a single block sticking out each side)
We know that Tn = n^2 + (n-1)^2 (as your son worked out)
so...
Total for shape n x n x n = (no. blocks in widest layer) + two layers either side of EACH diemnsions all the way down to one.
= (n^2 + (n-1)^2) + 2Σ( n^2 + (n-1)^2) [between n-1 and 1]
There is a formula for working out the sum of a recursive sequence above if I recall but I can't remember what it is.
I have probably missed the target by a mile and will be corrected by a 16 yr old but there we go... I will have forgotten more than he has ever known!
, but from a quick observation, the cubes are cumulative, i.e. the 1x1x1 fits inside and comprises the white cubes within the 2x2x2. Likewise the 2x2x2 shape makes up the white cubes in the 3x3x3 so the number of whites lags behind by one shape e.g.:shape, black, white, total
1x1x1, 1, 0, 1
2x2x2, 6, 1, 7
3x3x3,18, 6, 24
4x4x4, 38, 18, 56
You have the formula for the total in one layer (the widest layer in the shape that defines its dimensions), in order to get the whole shape you need to work it out thus:-
where Tn = total blocks in a layer of dimensions n x n
Total for shape = Tn + 2(Tn-1) + 2(Tn-2).... +2(T1) (to make up the shape there is the main layer in the middle and then 2 layers of each dimension, one on each side of the main layer for each dimension all the way down to one (every shape has a single block sticking out each side)
We know that Tn = n^2 + (n-1)^2 (as your son worked out)
so...
Total for shape n x n x n = (no. blocks in widest layer) + two layers either side of EACH diemnsions all the way down to one.
= (n^2 + (n-1)^2) + 2Σ( n^2 + (n-1)^2) [between n-1 and 1]
There is a formula for working out the sum of a recursive sequence above if I recall but I can't remember what it is.
I have probably missed the target by a mile and will be corrected by a 16 yr old but there we go... I will have forgotten more than he has ever known!
Last edited by ajm; 21 October 2004 at 12:07 AM.
21 October 2004, 01:40 PM
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How's this
So does a 4x4 shape contain 56(ajm) or 63(walwal) cubes
ajm the whites should equal the total of cubes at the previous level not just the blacks. So 63 is correct
As stated by ajm the white shapes fill the interior of the black shapes.
I visualize flattening the black shapes to 2D. So for 4x4 as stated the number of black cubes on the top is 4x4 + 3x3 = 25
The number of black cubes below the central plane is 3x3 + 2x2 = 13.
Blacks = 38.
So for a shape of dimension n the blacks = nxn + 2(n-1 x n-1) + (n-2 x n-2)
or n² + 2(n-1)² + (n-2)²
No time to go any further, soz
ajm the whites should equal the total of cubes at the previous level not just the blacks. So 63 is correct
As stated by ajm the white shapes fill the interior of the black shapes.
I visualize flattening the black shapes to 2D. So for 4x4 as stated the number of black cubes on the top is 4x4 + 3x3 = 25
The number of black cubes below the central plane is 3x3 + 2x2 = 13.
Blacks = 38.
So for a shape of dimension n the blacks = nxn + 2(n-1 x n-1) + (n-2 x n-2)
or n² + 2(n-1)² + (n-2)²
No time to go any further, soz
21 October 2004, 02:02 PM
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Look at the diagram and count the cubes. It's easier if you split it into layers and count the cubes in the layers. that ONLY gives you the number of the little cubes. If you're asking how many bigger cubes there are (e.g. ones made by joining others together, i.e. a 2x2 'theoretical' cube) then theres non in the 1x1x1, 2x2x2, 3x3x3 but there are 8 2x2 cubes and 1 3x3 cube in the 4x4x4 diagram.
21 October 2004, 03:50 PM
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Don't think that's the question.
Qu. Write formula for number of black (crust) cubes, and white (inner) cubes in the shapes, all cubes are 1x1x1, and hence a formula for the total number of cubes in the shapes. Awaits verification of this interpretation.
Can see where you're coming from but surely then you must have a greater number of cubes, not lesser? i.e. all the 1x1x1 plus your bigger cubes. There are 63 1x1x1 in the 4x4 shape.
Qu. Write formula for number of black (crust) cubes, and white (inner) cubes in the shapes, all cubes are 1x1x1, and hence a formula for the total number of cubes in the shapes. Awaits verification of this interpretation.
Can see where you're coming from but surely then you must have a greater number of cubes, not lesser? i.e. all the 1x1x1 plus your bigger cubes. There are 63 1x1x1 in the 4x4 shape.
21 October 2004, 04:05 PM
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Aha, I'm counting wrong! Ignore my comments on the 21 and 46 figures!!!
From the top layer
Layer 1 = 1 cube
Layer 2 = 5 cubes
Layer 3 = 13 cubes (for a while I only 'saw' 9 cubes!
)
Layer 4 = 25 cubes (again I was counting 16, silly me)
Layer 5 = 13 cubes
Layer 6 = 5 cubes
Layer 7 = 1 cube
Total = 63 cubes. (+ the other 'bigger' cubes I mentioned earlier).
The sequence seems to be
2x2x2 -1 = 7
3x3x3 -2 = 25
4x4x4 -1 = 63
so
5x5x5 -2 =123
6x6x6 -1 = 215
etc.
so nxnxn -2 (if n=odd num) or nxnxn -1 (if n=even) with the exception of 1x1x1 of course!
so, in theory 34x34x34 would have 39302 cubes.
Could be wrong as I'm thinking this up on the spot.
Can't help with the white vs black cubes bit but I'm sure someone can fit it in somewhere.
From the top layer
Layer 1 = 1 cube
Layer 2 = 5 cubes
Layer 3 = 13 cubes (for a while I only 'saw' 9 cubes!
)Layer 4 = 25 cubes (again I was counting 16, silly me)
Layer 5 = 13 cubes
Layer 6 = 5 cubes
Layer 7 = 1 cube
Total = 63 cubes. (+ the other 'bigger' cubes I mentioned earlier).
The sequence seems to be
2x2x2 -1 = 7
3x3x3 -2 = 25
4x4x4 -1 = 63
so
5x5x5 -2 =123
6x6x6 -1 = 215
etc.
so nxnxn -2 (if n=odd num) or nxnxn -1 (if n=even) with the exception of 1x1x1 of course!
so, in theory 34x34x34 would have 39302 cubes.
Could be wrong as I'm thinking this up on the spot.
Can't help with the white vs black cubes bit but I'm sure someone can fit it in somewhere.
21 October 2004, 04:42 PM
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LOL! Watch us make a meal of this!
Ok I added up the numbers of cubes wrong
, but my equation still stands:-
(1) Total Cubes for nxnxn shape = (n^2 + (n-1)^2) + 2Σ( n^2 + (n-1)^2) [ where Σ is evaluated between 1 and n-1]
(2) The number of white cubes is as above, but substitute in n-1 for n
(3) The number of black cubes is the (1) minus (2)
Ok I added up the numbers of cubes wrong
, but my equation still stands:-(1) Total Cubes for nxnxn shape = (n^2 + (n-1)^2) + 2Σ( n^2 + (n-1)^2) [ where Σ is evaluated between 1 and n-1]
(2) The number of white cubes is as above, but substitute in n-1 for n
(3) The number of black cubes is the (1) minus (2)
21 October 2004, 08:53 PM
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Hi guys... just got home from work... so soz not been back to you... this has certainly got the grey cells buzzing eh 
My son now has an extension to get this bit of homework... so will be taking a bit of time over the weekend which will surely be better than 1am last night
Thanks for your help so far... will get him to look at this and will post teacher's comments when *we* get the homework back
I still find it hard to believe that this is a GCSE question (EDEXEL)...
Many thanks again
My son now has an extension to get this bit of homework... so will be taking a bit of time over the weekend which will surely be better than 1am last night
Thanks for your help so far... will get him to look at this and will post teacher's comments when *we* get the homework back
I still find it hard to believe that this is a GCSE question (EDEXEL)...
Many thanks again
