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**Math Student****Guest**

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!

**Math Student****Guest**

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?

**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,908

1x1x24

1x2x12

1x3x8

1x4x6

2x2x6

2x3x4

6 combinations.

IPBLE: Increasing Performance By Lowering Expectations.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,908

56=7x8

1x1x56

1x7x8

2 combinations

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,908

100=5x5x4

1x1x100

1x4x25

1x5x20

4x5x5

4 combinations

IPBLE: Increasing Performance By Lowering Expectations.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,908

Let there be x cubes. the volume of each is 1.2^3=1.728 =>

x=216/1.728

x=125 cubes.

IPBLE: Increasing Performance By Lowering Expectations.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,908

you're absolutely right!

IPBLE: Increasing Performance By Lowering Expectations.

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

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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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,908

you're absolutey right!

IPBLE: Increasing Performance By Lowering Expectations.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,908

Can we find formula for this?

IPBLE: Increasing Performance By Lowering Expectations.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,908

If n is prime we have only 1 combination:

1x1xn

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,908

If n is product of two prime we have 2 combinations:

1x1xn and

1xp1xp2

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**Math Student****Guest**

Wow! Thank you for your help!

**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,908

I thought out an algoritm that can be used

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

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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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,908

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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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,908

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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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,908

Here is it in Mathematica language:

dq[x_] := Ceiling[Length[Divisors[x]]/2]

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,908

And here's a plot

*Last edited by krassi_holmz (2006-01-05 08:33:24)*

IPBLE: Increasing Performance By Lowering Expectations.

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