Why is $15m=0\pmod {18}$ equivalent to $3m=0 \pmod {18}$? $m$ is some natural number.  Why is $15m=0\pmod {18}$ equivalent to $3m=0 \pmod {18}$?
This is my question basically. I don't understand why that's true.
 A: Because you get one equation by multiplying the other by $5$, and $5$ is invertible modulo $18$ (namely, $5\cdot 11\equiv 1 \pmod{18}$).
So if $3m\equiv 0\pmod{18}$ then also $5\cdot 3m\equiv 5\cdot 0\pmod{18}$ which is the same as $15m\equiv 0\pmod{18}$. And if $15m\equiv 0\pmod{18}$ then also $11\cdot 15m\equiv 11\cdot 0\pmod{18}$ which is the same as $3m\equiv 0\pmod{18}$.
A: The statement:
$$15m\equiv 0 \bmod 18$$
means that 18 divides $15m$, i.e., there's an integer number $k$ with
$$
18k = 15m. 
$$
(That's straight from the definition of modular equivalence.)
If you subtract $18m$ from both sides, you get
$$
18(k-m) = -3m
$$
and if you then negate, you get
$$
18(m-k) = 3m
$$
That shows that $3m$ is also divisible by 18, which is the same as 
$$
3m \equiv 0 \bmod 18.$$ 
When you really understand what I've written, look at @HenningMakholm's
answer for possible deeper insight. 
A: We have to show that $18|15m\iff 18|3m$
Clearly if $18|3m$ then $18|5(3m)$
The hard part is showing that if $18|15m$ then $18|3m$.
Suppose that $18|15m$, then there exists an integer $k$ with $5\times 3\times m=18\times k$.
This implies that $5$ divides $18\times k$, since $5$ is a prime we conclude $5$ divides $k$ by Euclid's lemma, so $k=5j$.
Finally notice $5\times3\times m=18\times 5\times 5\implies 3m=18j\implies 18|3m$
