Congruences - Change of Modulo Base Let's say I have the following congruence:
x ≡  y (mod b)
Is there a formular, theorem, or algorithm that gives the new (congruent) relationship between x and y when I change the modulus base b to something else.
Change to something else could be any arithmetic operation or combination of operations performed on b.
 A: Only if what you change to is a divisor of $b$. In that case it's straight-forward. For instance, if we have
$$
x\equiv 7\pmod{15}
$$
we can easily see that
$$
x\equiv 2\pmod 5
$$
However, other than bases $5$ and $3$ (and possibly $1$ if you count that), there is no other modular base where we know exactly what $x$ is.
For multiples of $15$, we can rule out a lot of options, for instance we know that
$$
x\equiv 7\text{ or }22\pmod{30}
$$
and for numbers that share factors with $15$, we can similarily rule out a few options. For instance, we must have
$$
x\equiv 2\text{ or }7\pmod{10}
$$
because of the shared factor $5$ between $10$ and $15$. Basically, if we reduce to modulo $5$, the two must give the same class for $x$, and the two classes modulo $10$ that give $2$ modulo $5$ are $2$ and $7$.
If you know $x$ in other modular bases, those could be combined using the Chinese remainder theorem, as long as they're not contradictory ($x\equiv 3\pmod{10}$ would, for instance, be contradictory to the above example). For instance if we also know that $x\equiv 0\pmod 2$, then we necessarily have
$$
x\equiv 22\pmod{30}
$$
A: Notice that $x\equiv y \mod b$ is equivalent to say that $\exists k\in \mathbb Z$ such that $x-y=kb$. 
Then if you take any $q|b$ then $x-y=kq\bar b$ so $$x\equiv y\mod q$$
