Mathematics

So, how are you people doing with the two problems I gave you? Like it? HAHAHA. Alright, to those who had some hard time with the two questions, hold your breath! There's nothing bout that today. Hee. :p Sorry for those who are waiting for the third part. I am not going to update this series now. Thanks to my tutor who got me into this math thing again. Proving something really interesting. 

I really want to share this with you but not now. He gave us 3 maths problems to prove. However, I only managed to prove 2 of them, which according to him, that's SUPER EASY and Slightly easy. I am yet to prove the SIPEH (super, incredibly, riddiculusly _____(insert whatever you like) HARD proof. I already spent my whole afternoon trying to prove it. However, still I failed. Damn! 

Some of you might be wondering why I am so obsesed with it? After all, it's just a math problem. Honestly, this is not about math now. I admit I am really facinated by numbers and math, however, sometimes pride is the issue also. He told us a story behind this proof. I will tell you the whole story after I figure out the proof. I will only tell you how lame you are if you are unable to proof the question that I am going to post later in this post. This math problem is being proved decades ago by a BOY! He's only 7 year old then. Now? Harlo! How old are you now? If you can't prove this, you lost to a kid man. LOSER! (Well, that's what kept me cracking my head for this afternoon.) However, I solved it in the end. It's this another problem that makes me go cockoo whole afternoon! 

Nevermind bout that, just to make some noise bout losing to a kid. Hee. So here are the questions. I will post the answer some other time.

Let S = { 1, 2, 3, ... , 100}

n(A) = 51

A is a subset of S

I know some of you are thinking what the hell is this?

let me explain. S is a set which contains all the numbers from 1 to 100.

now 51 numbers were chosen randomly. 

Alright?

Now prove that 

i) There exist 2 numbers in the 51 chosen numbers such that the sum of these 2 numbers is 101.

ii) There exist 2 numbers in the 51 chosen numbers such that the Highest Common Factor (Greatest Common Divisor) of these 2 numbers is 1. 

*this is what the KID proved*

iii) There exist 2 numbers in the 51 chosen numbers such that one of the number can divide the other number completely. 

*this is where I got stuck*

So enjoy proving this. Muahahahaha.

for now, I will continue to crack my head.

- =.= -