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.