Prove the following using fundamental definitions.
Suppose $a$, $b$, and $c$ are positive integers such that $a|b$ and $a|c$. Then for any integer $k$, $ka + 3b ≡ c \mod a)$
What I've done:
"We need to show $ka + 3b ≡ c(mod a)$. So by the definition of congruent we need to show $ka + 3b - c = am$ with $m$ being an integer. So, we need to show $a|(ka + 3b -c)$.
Since $a|b$ then $a|3b$. Since $a|c$ then $a|-c$. Also, note $a|a$ so $a|ka$ for any integer $k$
Notice then that $a|(ka + 3b -c)$ since $a>0$
Is this a valid proof?
Notice that since $a\mid c$, $$c\equiv 0\pmod a.$$
Now, since $a\mid b$, then $$ka+3b\equiv 0\pmod a.$$
The claim follow.