Find the indicated orders: Find the indicated orders below:
$$(a) ord_{8}3$$
$$(b) ord_{11}3$$
$$(c)ord_{20}3$$
$$(d)ord_{20}19$$
$$(e)ord_{35}1$$
I'm not sure what this is asking for.  If someone could either give me a hint, or possibly work out one or two to show me how I can finish the rest, it would be great.
Thanks,
Chloe :)
 A: The number $\text{ord}_m(a)$ is the smallest positive integer $e$ (if it exists) such that $a^e\equiv 1\pmod{m}$.
If $a$ and $m$ are relatively prime, there are $k\gt 0$ such that $a^k\equiv 1\pmod{m}$, otherwise there are not. 
In principle, to find the order of $a$, where $\gcd(a,m)=1$, we can try all $k\ge 1$ until we bump into the answer. For small $m$, it will not take terribly long, since the order of $a$ divides $\varphi(m)$, where $\varphi$ is the Euler $\varphi$-function.
In practice, we will want to take shortcuts. For example, consider the order of $99$ modulo $100$. Since $99$ is congruent to $-1$, we have $99^2 \equiv (-1)^2 \pmod{100}$, so the order of $99$ is $2$ or $1$, but it is obviously not $1$. 
The fact that the order of $a$ modulo $m$ divides $\varphi(m)$ can be useful. For example, suppose we want to find the order of $7$ modulo $40$. Since $\varphi(40)=16$, the only candidates for the order of $7$ are $1$ (ridiculous), $2$ (clearly not), $4$, $8$, or $16$. It is relatively easy by repeated squaring to compute the appropriate powers modulo $40$. 
There are a few other tricks. Your particular examples are computationally not hard. I will stay away from them, but mention a couple of related examples.
What is the order of $4$ modulo $15$? Certainly not $1$! But $4^2\equiv 1\pmod{15}$, so the order is $2$.
What is the order of $3$ modulo $13$? Note that $3^3\equiv 1\pmod{13}$, and no power smaller than the cube will work, so the order of $3$ modulo $13$ is $3$. 
Of your questions, by far the easiest is (e).
