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

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

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]

56 and 100 both have 4 prime factors, but they give answers of 6 and 8.

]]>1x1xn and

1xp1xp2]]>

1x1xn]]>

56 = 2x2x2x7

1x1x56

1x2x28

1x4x14

1x7x8

2x2x28

2x4x7

6 combinations.

100 = 2x2x5x5

1x1x100

1x2x50

1x4x25

1x5x20

1x10x10

2x2x25

2x5x10

4x5x5

8 combinations.

]]>x=216/1.728

x=125 cubes.]]>

1x1x100

1x4x25

1x5x20

4x5x5

4 combinations

]]>