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#1 2005-10-29 06:03:35
- robin
- Novice
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Rectangular arrangements
Whilst tiling the kitchen, I got to thininking about what shapes could be made with the tiles. in particular rectangles.
For example if I had 15 tiles, how many different arrangements of rectangles could I make. In this case it was two.
15 tiles = 2 arrangements 1 x 15 and 3 x 5
A stack of 1111 walls tiles ???
How many rectangular arrangements can be maded using 1111 tiles. Is there a formular for working this out ?
Ive searched for an answer on the net, but haven't found anything to help.
Any pointers would be great
Robin
#2 2005-10-29 07:37:52
- MathsIsFun
- Administrator
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Re: Rectangular arrangements
This may help: Prime Factorization Tool
But you may be thinking "What what what?"
Well, it doesn't exactly answer your question, but it gives you the elements you need to get your answer.
For 1111 it gives 11 × 101, so there aren't many combinations there.
But for 100 it gives 100 = 2² × 5², which means 2 × 2 × 5 × 5, so you could use combinations of those to figure out:
1 × 100
2 × 50
4 × 25
5 × 20
10 × 10
Have a play and tell me what you find out. 
"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman
#3 2005-10-29 08:01:28
- mathsyperson
- Moderator
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Re: Rectangular arrangements
I had a version of that ready to post, and then my internet broke temporarily. Grr.
Your version's better, so it's probably for the best, but still... grr.
Edit: I got thinking about it for a bit, and I believe that if a number splits into prime factors of the form a^b*c^d*e^f*..., then it will have in total (b+1)*(d+1)*(f+1)*... factors. As each of these factors has a partner that it can multiply with to produce the original, then you would just divide the result of the above formula by 2 to find the number of arrangements. The exception to this is when the original number is a square, because then one of its factors would group with itself to produce the original. Luckily, square numbers, and only square numbers, produce an odd number of factors, so when you try to find the amount of arrangements using the above method you will get a remainder and so when this happens you will know to just round up.
Disclaimer: All of that stuff was just from me thinking and I didn't actually calculate anything, so I can't vouch for its accuracy. I think its right though.
Why did the vector cross the road?
It wanted to be normal.