Recent content by sagardipak

The main reason for 0!=1 is that it fits many different situations, like the ones that everyone posted above. But actually, the factorial is a function that is defined as: n! = \prod_{k=1}^{n}k, \forall n \in \mathbb{N} (in an informal language: n! = 1 x 2 x ... x n) Since the domain of...

First, we have to understand that the absolute value is a function defined by: |x| = \begin{cases} x & \text{if } x\geq 0 \\ -x & \text{if } x<0 \end{cases} So, |ab| = \begin{cases} ab & \text{if } ab\geq 0 \\ -ab & \text{if } ab<0 \end{cases} Now, let's see what |a||b| is...

Are you saying that if n is an odd perfect square, then n divides a number of the form 10^k+1? If so, that is not correct. Take n=9. It will never divide any number of that form. Actually, the division will always have a remainder of 2. Check that ;)

Well, answering your question, the 11 plays a special role here. You can easily prove that 10^k+1 is a multiple of 11 for every odd k. Are you familiar with modular arithmetic? It is very helpful in number theory.