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Please help me on this question!
24 unit cubes can be stuck together to make cuboids of different shapes. How many DIFFERENT cuboids can be made?
How many different cuboids can be made with:
a) 56 cubes
b) 100 cubes
Thanks In Advance!
In fact, there is another question:
Many small cubes of side 1.2 are stuck together to make a large cube of 216cm³.
How many cubes are needed.
I worked it out to be 125 but can anyone confirm that answer?
1x1x24
1x2x12
1x3x8
1x4x6
2x2x6
2x3x4
6 combinations.
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56=7x8
1x1x56
1x7x8
2 combinations
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100=5x5x4
1x1x100
1x4x25
1x5x20
4x5x5
4 combinations
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Let there be x cubes. the volume of each is 1.2^3=1.728 =>
x=216/1.728
x=125 cubes.
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you're absolutely right!
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There are a few more combinations for 56 and 100, because 8 and 4 can be broken down more.
56 = 2x2x2x7
1x1x56
1x2x28
1x4x14
1x7x8
2x2x28
2x4x7
6 combinations.
100 = 2x2x5x5
1x1x100
1x2x50
1x4x25
1x5x20
1x10x10
2x2x25
2x5x10
4x5x5
8 combinations.
Why did the vector cross the road?
It wanted to be normal.
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you're absolutey right!
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Can we find formula for this?
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If n is prime we have only 1 combination:
1x1xn
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If n is product of two prime we have 2 combinations:
1x1xn and
1xp1xp2
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Wow! Thank you for your help!
I thought out an algoritm that can be used
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Be careful, though. As we can see from the example above, how many prime factors a number has isn't the only thing it depends on.
56 and 100 both have 4 prime factors, but they give answers of 6 and 8.
Why did the vector cross the road?
It wanted to be normal.
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Yes, I saw that. My first result:
Ler dq[x] gives the number of the different divisors of x. Then the number of diffrerent rectangles that can be made with n squares is
Ceiling[dq[x]/2]
Last edited by krassi_holmz (2006-01-05 08:16:23)
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Example:
number:36
divisors:1,2,3,4,6,8,9,12,18,36
dq[36]=10
So we must have 5 rectangles. Here are they:
1x36
2x18
3x12
4x9
6x6
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Here is it in Mathematica language:
dq[x_] := Ceiling[Length[Divisors[x]]/2]
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And here's a plot
Last edited by krassi_holmz (2006-01-05 08:33:24)
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