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        riddles >> putnam exam (pure math) >> Re: Easy math Q. for you guys. Hard for me.
        
(Message started by: towr on Dec 10^th, 2006, 9:36am)

Title: Re: Easy math Q. for you guys. Hard for me.
Post by towr on Dec 10^th, 2006, 9:36am

I think you'd be better off asking in the other thread. People will most likely not consider it pestering.

Also, it would be quite good to have the actual topic of a thread mentioned or hinted at in some way in the title.

Title: Re: Easy math Q. for you guys. Hard for me.
Post by towr on Dec 10^th, 2006, 9:52am

Ok, from the other thread (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_hard;action=display;num=1165677723) I gather you are given a number X, and want to cut a square into X smaller squares. (This is an easier way to look at the problem.)

One obvious way is to cut a square into 4 squares (one 2x2 square -> four 1x1 squares).
You can repeat this, buy cutting any of the smaller squares into 4 again
(one 4x4 square -> four 2x2 -> three 2x2 square and four 1x1 squares)
Using this method, you always divide a square in 1+3k squares (1,4, 7, 10, 13 etc)
Every time you cut a squre in 4, you increase the number of squares by 3 (you remove a big one, and add 4 smaller ones; 4-1=3)

Another option, is to cut a square into 9 squares
(one 3x3 square -> nine 1x1 squares)
Using just this method, you can get any number 1+8k (1,9,17, 25, 33, etc)
It's similar to before; you destroy a big square, and create 9 new smaller ones from it. So the net increase is 8.

Of course, you can also combine the two methods of dividing a square into squares. This means you can get any number that may be represented as 1 + 3k + 8m (1, 4, 7, 9, 12, 15, 17, 18, 20, etc)

From some point onwards, you can get any X that is greater.

Title: Re: Easy math Q. for you guys. Hard for me.
Post by towr on Dec 10^th, 2006, 10:29am

on 12/10/06 at 10:07:19, koonie32 wrote:

Hm ok.
So how does this work for say x = 6, 8, or anything other than (1,4,7, etc.). I mean or 6 you could do 1 4x4 and 5 2x2. This is not in the formula.. or is it?

Ah, very well spotted. I missed a few cases.
For bigger X it wouldn't matter, because you'd be able to get them all. But you're quite right.

So if you have a 3x3 square (or multiple thereof), you can cut it in 6 pieces. So using just that, you can get any number 1+5k

Together with the other two, we can now get 1+3k+ 8m + 5l.
Of course. 8 = 3+5, so the previous formula is equal to 1+ 3(k+m) + 5(l+m).
So this means we can use your method instead of my second one: cut squares into 4 (adding 3 to the total) or 6 (adding +5)
This gives 1,4,6,7,9,10,11,12,14,15,16 etc (barring mistakes)

I don't yet see a way to get 2,3,5,8 and 13 though.

[edit]We can get 8 with 3x3 + 7 * 1x1. And from that, by adding 5, we also get 13
So only 2,3 and 5 we can't get[/edit]