Question

How many positive integers $n$ satisfy $127 \equiv 7 \pmod{n}$? $n=1$ is allowed.

Answer 1

Naturally, any integer $n$ larger than 127 will return $127\equiv127\mod n$, and of course $127\equiv0\mod127$, so we restrict the possible solutions to $1\le n<127$.

Now,

$127\equiv7\mod n$

is the same as saying there exists some integer $k$ such that

$127=nk+7$

We have

$\implies 120=nk$

which means that any $n$ that satisfies the modular equivalence must be a divisor of 120, of which there are 16: $\{1,2,3,4,5,6,8,10,12,15,20,24,30,40,60,120\}$.

In the cases where the modulus is smaller than the remainder 7, we can see that the equivalence still holds. For instance,

$127=21\cdot6+1\iff127\equiv1\equiv7\mod6$

(If we're allowing $n=1$, then I see no reason we shouldn't also allow 2, 3, 4, 5, 6.)