Why is (-3/77) mod 65 equal to 16? Getting an X for Chinese Remainder Theorem (CRT)
In the "Easy CRT" part of the answer to this problem, the author demonstrates that (-3/77) mod 65 is equal to 16. I don't understand - how is this accurate? I sort of understand the steps, but wouldn't the answer just be 62/77?
Thanks, and I apologize if I've missed something obvious!
 A: The definition of $\frac{1}{x}$ is that $\frac{1}{x}$ is the quantity such that $x \cdot \frac1x = 1$ (which may or may not exist). Therefore
$$ \frac{-3}{77} \equiv 16 \pmod {65} \text{ if and only if } -3 \equiv 77 \cdot 16 \pmod {65} $$
This happens if and only if
$$77 \cdot 16 + 3\equiv 0 \pmod {65}$$
which by definition of congruence mod $65$ says that
$$65 \mid (77 \cdot 16 + 3)$$
which is true because $$ 77 \cdot 16 + 3 = 1235 = 65 \cdot 19. $$
A: In the ring of integers mod. $n$, there is no ‘real’ fraction. Some elements have reciprocals, others  don't, depending whether they're coprime with the modulus or not.
For instance, modulo $14$, $5$  is a unit, since $3\cdot 5\equiv 1\pmod{14}$, and $5^{-1}=3$, so one can be tempted to write, say, $\dfrac 45$ instead of $4\cdot 5^{-1}=4\cdot 3=12\pmod{14}$. This way of writing is very confusional, from my point of view.
In the present case, we have $77\equiv 12\pmod{65}$ and one computes with the extended Euclidean algorithm that $12^{-1}\equiv -27$, so that
$$-\frac 3{77}=(-3)(-27)=81\equiv 16\pmod{65}.$$
