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krassi_holmz
2006-01-06 07:32:34

And here's a plot

krassi_holmz
2006-01-06 07:27:14

Here is it in Mathematica language:
dq[x_] := Ceiling[Length[Divisors[x]]/2]

krassi_holmz
2006-01-06 07:15:16

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

krassi_holmz
2006-01-06 07:12:12

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]

mathsyperson
2006-01-06 07:03:21

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.

krassi_holmz
2006-01-06 06:43:44

I thought out an algoritm that can be used

Math Student
2006-01-06 06:42:45

Wow! Thank you for your help!

krassi_holmz
2006-01-06 06:41:24

If n is product of two prime we have 2 combinations:
1x1xn and
1xp1xp2

krassi_holmz
2006-01-06 06:39:33

If n is prime we have only 1 combination:
1x1xn

krassi_holmz
2006-01-06 06:38:31

Can we find formula for this?

krassi_holmz
2006-01-06 06:37:17

you're absolutey right!

mathsyperson
2006-01-06 06:26:13

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.

krassi_holmz
2006-01-06 06:14:22

you're absolutely right!

krassi_holmz
2006-01-06 06:08:55

Let there be x cubes. the volume of each is 1.2^3=1.728 =>
x=216/1.728
x=125 cubes.

krassi_holmz
2006-01-06 06:04:29

100=5x5x4

1x1x100
1x4x25
1x5x20
4x5x5

4 combinations