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        riddles >> hard >> Re: Another hard math question
        
(Message started by: THUDandBLUNDER on Dec 9^th, 2006, 8:03am)

Title: Re: Another hard math question
Post by THUDandBLUNDER on Dec 9^th, 2006, 8:03am

So X can be 4? Can it be 9?
How do you arrange 4 or 9 tiles into a square?

Where do you get all these hard maths questions from? Homework? What grade are you in?
(Please put them in Easy in future as once you know the answer they become easy, right?)

Title: Re: Another hard math question
Post by THUDandBLUNDER on Dec 9^th, 2006, 8:45am

on 12/09/06 at 08:22:54, koonie32 wrote:

Duh I got the squares one easy (4,9, etc.).
But what about 6,7,8,10, etc.
I mean I can draw them say with 7

Do 3 4X4 and 4 2x2

But is there a formula for N and not a specific number for N?

Do you have an unlimited amount of smaller tiles of size 1,2,3,4....etc?
What is N? There can't be a formula without a definition. Is it the size of the square you are trying to form?
What is 'X'? What is 'x'?

If you want help it is up to you to first clearly explain to others what you are trying do.

Title: Re: Another hard math question
Post by Whiskey Tango Foxtrot on Dec 9^th, 2006, 8:59am

Let me try to get this straight. You're tiling a square of a given size, say unit length sides. You can change the size and number of the tiles you're putting on it with the exception that the number cannot be 2,3, or 5. And you want an expression for the number of tiles needed? A minimum number of tiles or just any number?

Title: Re: Another hard math question
Post by THUDandBLUNDER on Dec 9^th, 2006, 9:18am

on 12/09/06 at 08:55:13, koonie32 wrote:

Ok
N is the number of squares that you can have. The squares can be of different sizes.

OK, give this to your teacher: ;D

Title: Re: Another hard math question
Post by THUDandBLUNDER on Dec 9^th, 2006, 9:54am

Continuing your example with 7 tiles, adding 3 more 4x4 tiles to the 4x4 square makes an 8x8 square.
Adding 3 more 8x8 tiles to the 8x8 square makes a 16x16 square, etc.

So I think the formula you are looking for is
N = 3n - 2

Title: Re: Another hard math question
Post by THUDandBLUNDER on Dec 9^th, 2006, 10:57am

Quote:

There must be a way to tile a square with x other squares with a formula.

I gave you a formula.
When x = 3n-2 (for n = 3,4,5.....) you can make a larger square.
So it is possible with 7, 10, 13, 16, 19 etc. tiles.
And when 3n-2 smaller squares are used the side of the larger square formed is 2^n-1 long.

Title: Re: Another hard math question
Post by THUDandBLUNDER on Dec 9^th, 2006, 11:18am

on 12/09/06 at 11:10:11, koonie32 wrote:

Wait but in your example are x and n the same thing? Or is N equal to a side of one of N squares?

Look, what you should do is take my 'x', multiply by the square root of your 'X', divide by my 'n', raise all of that to the power of your 'N', if it is Saturday take away three times the number you first thought of, and that's your answer except when there is a full moon, OK?

on 12/09/06 at 11:10:11, koonie32 wrote:

I just realized there was a "pure math" place. I should of posted it there. Sorry.

Good luck to them.

Title: Re: Another hard math question
Post by Sameer on Dec 9^th, 2006, 1:33pm

on 12/09/06 at 11:10:11, koonie32 wrote:

Wait but in your example are x and n the same thing? Or is N equal to a side of one of N squares?

I just realized there was a "pure math" place. I should of posted it there. Sorry.

Given the complexity of the math you are doing, it looks like high school math. Post it in easy. And use some relevant titles instead of "hard math question" ...

A lot of people here are way past their colleges. Lot of them hold degress in Math. Lot of them are engineers and Lot of them are college students. So you can compare the difficulty levels now...

Are you just not asking us to do your homework now is it? ::)

Title: Re: Another hard math question
Post by markr on Dec 10^th, 2006, 11:19pm

If you can do it with N squares, you can do it with N+3 squares (divide one of the N squares into four smaller squares). Therefore, if it can be shown possible for three consecutive integers (N, N+1, N+2), then it is possible for all integers greater than or equal to N.

The only positive integers it can't be done with are:
2, 3, 5, 6, 8, 11, and 14. Therefore, N=15.

15: 8 1x1 (=2 2x2), 7 2x2
16: 16 1x1
17: 9 1x1 (=1 3x3), 8 3x3

Title: Re: Another hard math question
Post by towr on Dec 11^th, 2006, 1:12am

on 12/10/06 at 23:19:06, markr wrote:

It can be done for 6, 8 and thus 11 and 14
2x2 + 5 * 1x1
3x3 + 7 * 1x1

Title: Re: Another hard math question
Post by jollytall on Dec 11^th, 2006, 6:22am

So it seems that the original riddle was the solution. Or with otherwise it could have worded, that "Prove that for any x<>2, 3, 5 a squere can be covered with x smaller squeres."