Congruence of Quotient

Theorem
Let $$a, b \in \Z$$ and $$n \in \N$$.

Let $$a$$ be congruent to $b$ modulo $n$, i.e. $$a \equiv b \left({\bmod\, n}\right)$$.

Let $$d \in \Z: d > 0$$ such that $$d$$ is a common divisor of $$a, b$$ and $$n$$.

Then $$\frac a d \equiv \frac b d \left({\bmod\, \frac n d}\right)$$.

Proof
By definition of congruence modulo $n$:
 * $$a = b + k n$$

Dividing through by $$d$$ (which you can do because $$d$$ divides all three terms), we get:
 * $$\frac a d = \frac b d + k \frac n d$$

from where the result follows directly.

Alternative Proof
From Congruence by Product of Modulo, we have that:
 * $$\frac a d \equiv \frac b d \left({\bmod\, \frac n d}\right) \iff d \frac a d \equiv d \frac b d \left({\bmod\, d \frac n d}\right) \iff a \equiv b \left({\bmod\, n}\right)$$