Why is finding factors until half of the number enough? If I want to find factors of a number except itself, at first I think that I divide it the number in turn $1, 2, 3, ... , n - 1$. However, after a while, noticed that division to get the factors is sufficient up to $n/2$(inclusive). The rest part is not needed. What is its reason? Why is it enough to find them? How can it be explained?
@Edit, to find prime number until $\sqrt n$ is sufficient, but what about perfect numbers? I think $n/2$ is good way to go perfect numbers?
For 6,
1   2   3   4   5
^
    ^
        ^|
           Up to here, it's ok.


Actually to find the (prime) factors of a whole number $n$, it is enough to check for divisors up to $\sqrt n$.  Going up to $n/2$ would be overkill.
One way to think about this is to ask, if $n$ is not a prime but instead composite, how big can the smallest divisor of $n$ be?  If we have the factorization $n = a\cdot b$, we cannot have both $a$ and $b$ larger than $\sqrt n$.
To make a complete list of the factors of a whole number $n\gt 1$, you can do it by checking all $1\lt a \lt \sqrt n$ and putting both $a$ and $b$ in your list when the division $n/a = b$ is exact.  If $\sqrt n$ happens to be an integer (i.e. $n$ turns out to be a perfect square when we extract $\sqrt n$), then put one copy of that integer in the list of divisors. 
 The list (of proper divisors of $n$) will be complete when you finish by including $1$, which is always a divisor you want in your list.
