1. ## Quick question...

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

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

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

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

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

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