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30 M Azores
speaks English and Portuguese and Saraiki
Last login: 24 August 2011
KupiCredit: 0
Sent comments: 9
Received comments: 0
Profile views this month: 0
Member since: 14 January 2011
 "A mathematician is a blind man in a dark room looking for a black cat which isn't there." - Charles Darwin
I know that you believe you understand what you think I said,
but I'm not sure you realize that what you heard is not what I

About -fake made up name here-.

I can be a very nice person, if you let me. Boring, humdrum people bore me to death, it's just so fake now. Be a little lively and yourself, will you? Don't need to be fake here, even though that's why I don't converse with new members anymore because I know they played Kupika before and are just fakers or another millionth multiple account that could be talking to me with another of their accounts but act like different people(happened already to me) Thanx IP Ad. and uncle who can crack shit like that. Thank you and good day. If I don't reply, I find you as one of them. Change my mind, please.

And ha ha ha. 1,2,4,?,16,32,64? Hm no. It should be more like, let's see. Oh jesss! Show that the number of squares whose vertices lie at the centers of the squares of an n by n checkerboard is given by n2(n2 - 1)/12. As an example, for the case 4 by 4, there are: * 9 squares of size 2 x 2 (like the green one) * 4 squares of size 3 x 3 (like the blue one) * 1 square of size 4 x 4 (like the red one) * 4 diamonds (like the yellow one) * 2 skewed squares (like the purple one) for a total of 20. Solution If we consider the number of squares that have their vertices on the border cells of an k by k grid, then the number is clearly (k-1). For example, there are three such squares on a 4 by 4 grid: Within an n x n grid, we have: * (n-1) squares based on the border of the n x n grid itself * (n-2) squares based on the border of each (n-1) by (n-1) subgrids, of which there are 4 * (n-3) squares based on the border of each (n-2) by (n-2) subgrids, of which there are 9 * ... * 1 square based on the border of each 2 by 2 subgrid, of which there are (n-1)2 The total number of squares is thus: (n-1)12 + (n-2)22 + ... + 2(n-2)2 + 1(n-1)2 = Σ k=1,2..(n-1) (k(n-k)2) = Σ k=1,2..(n-1) ((n-k)k2) = Σ k=1,2..(n-1) (nk2 - k3) = n[n(n-1)(2n-1)/6] - n2(n-1)2/4 = n2(n-1)[(2n-1)/6 - (n-1)/4] = n2(n-1)(n+1)/12 = n2(n2-1)/12 And that is just some 9th grade Geometry, good lord! Simple. XDD I make myself laugh. And the numbers would change some every day, giving a different answer. Maybe it being 9th grade is too kind? If this was the case, nobody would feel like making new accounts. Ha ha ha. I mean, would you go that far? Maybe for your friends here, yes, but to be fake? Nah. TROLOLAALA.


Q&A Section   
‹colours and carousels♥› 21 May 11  

The maths on your profile makes my brain hurt.
KupiciouS 13 Jul 11  
Late reply is late.

Poor brain, sorry about that.
‹N๏xx› 15 Jan 11  
It didn't sound that rude at all, don't worry.
I fell asleep anyway.
Surely that's the pinnacle of rudeness?
KupiciouS 15 Jan 11  
I'm alright with it. I had to leave anyway.
‹N๏xx› 15 Jan 11  
If you're sure. Are you up to much?
KupiciouS 15 Jan 11  
Today, not much. I'm having guests coming over in half an hour and the menudo
is not ready yet. I'm going to be getting off soon. You'll know when I don't
reply. Sorry, this sounds so rude.
‹N๏xx› 14 Jan 11  
Thank you ~
It's nice to meet you, by the way, forgive my lack of manners.
KupiciouS 15 Jan 11  
Nice to meet you too. It's fine, it's fine. Being etiquette...I don't have much
of it.
‹N๏xx› 14 Jan 11  
No need to apologise, I was curious is all. 
Choose a random number between 1 and 50 for me, would you, m'dear?
KupiciouS 14 Jan 11  
Forty-eight. (:
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