<s>[INST] finding a primitive root. It says for part A to Find a primitive root r of 38? Im not sure if I did it right. 
I first calculated $\phi(38)=\phi(19*2)=18$. So there are 18 numbers that are relatively prime to 38. Listing them out we get 1,3,5,7,9,11,13,15,17,21,... so on. So I decided to test out 1.
But $ord_1 38$  does not equal $\phi(38)$. But $ord_3 38=\phi(38)$. 
So I calculated it until i found a power of 3 that was congruent to $1 \pmod {38}$ 
So I get $$3^1≡ 3 \pmod {38}$$ $$3^2 ≡ 9 \pmod {38}$$ $$3^3 ≡  27 \pmod {38}$$  $$3^4 ≡ 5 \pmod {38}$$ $$ 3^5  ≡ 15 \pmod {38}$$ $$3^6  ≡ 7\pmod {38}$$ $$3^7  ≡ 21 \pmod {38}$$ 
$$3^8 ≡ 25 \pmod {38}$$ $$3^9  ≡ -1 \pmod {38}$$  $$3^{16}  ≡ 17 \pmod {38}$$ $$3^{17}  ≡ 13 \pmod {38}$$ $$3^{18} ≡ 1 \pmod {38}$$  So $ord_3 38=18=\phi(38)$.  
 [/INST] As $\phi(19)=18$ and $ord_{19}2$ must divide $\phi(19),$ we only need to test for the powers $1,2,3,6,9,18$
$3^2=9;3^3=27\equiv-11\pmod{38};3^6\equiv(-11)^2\equiv121\equiv7\pmod{38}$
So, $3^9=3^3\cdot3^6\equiv(-11)(7)\equiv-77\equiv-1\pmod{38}$
$\implies 3 $ is a primitive root $\pmod{38}$

Alternatively, let's start with finding a primitive root of $19$
Starting with $2$, the smallest positive integer $>1$
$2^2=4,2^3=8,2^6=64\equiv7\pmod{19},$
$2^9=2^3\cdot2^6\equiv 8\cdot7\pmod{19}\equiv-1$
$\implies 2 $ is a primitive root $\pmod{19}$
Now, as $(2,38)=2>1,2$ can not be a primitive root $\pmod {38}$
But, $ord_{19}(2+19k)=ord_{19}2=18$
Now, $ord_2(2+19k)$ will be $1$ if $(2+19k)$ is odd i..e, if $k$ is odd
So, $ord_{(2\cdot19)}(2+19k)=$lcm$(1,18)=18=\phi(38)$ if $k$ is odd
Putting $k=2r+1$ where $r$ is any integer, $2+19k=2+19(2r+1)\equiv21\pmod{38}$
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