1. Quick question...

Lets say you have a number like $\displaystyle 2^{1000}$. Then the last two digits of this number is the remainder of $\displaystyle \frac{2^{1000}}{100}$ in $\displaystyle \mathbb{Z}/100 \mathbb{Z}$. And $\displaystyle \mathbb{Z}/100 \mathbb{Z}$ contains $\displaystyle 100$ residue classes.

Is it okay to generalize, and say that the last $\displaystyle n$ digits of a number $\displaystyle k^{p}$ where $\displaystyle k > 0$ and $\displaystyle n \leq p$ is the remainder of $\displaystyle \frac{k^{p}}{n}$ in $\displaystyle \mathbb{Z}/n \mathbb{Z}$? And the last $\displaystyle n$ digits are in one of the $\displaystyle n$ residue classes?

2. Hello,
Originally Posted by particlejohn
Lets say you have a number like $\displaystyle 2^{1000}$. Then the last two digits of this number is the remainder of $\displaystyle \frac{2^{1000}}{100}$ in $\displaystyle \mathbb{Z}/100 \mathbb{Z}$. And $\displaystyle \mathbb{Z}/100 \mathbb{Z}$ contains $\displaystyle 100$ residue classes.
Yes

Is it okay to generalize, and say that the last $\displaystyle n$ digits of a number $\displaystyle k^{p}$ where $\displaystyle k > 0$ and $\displaystyle n \leq p$ is the remainder of $\displaystyle \frac{k^{p}}{n}$ in $\displaystyle \mathbb{Z}/n \mathbb{Z}$? And the last $\displaystyle n$ digits are in one of the $\displaystyle n$ residue classes?
Huh ?
The last n digits are the remainder in the Euclidian division of $\displaystyle k^p$ by $\displaystyle {10^n}$

What I don't understand is why you bother yourself talking about residue classes... ?

3. thats what I meant divide by $\displaystyle 10^{n}$. Because the remainder is a residue class.