1. ## More congruence trouble

for which integers $c$ with $0 \leq c < 1001$ does the congruence $154x \equiv c$ $(\bmod 1001)$ have solutions? For each case when there are solutions, find how many solutions are not congruent modulo 1001.

Originally, I thought this problem might have something to do with multiplicative inverses, but I didn't get anywhere on that front.

If it matters, I calculated $\gcd (154,1001)$ to be 77.

2. $ax \equiv b \mod n$ has a solution iff $(a,n) \mid b$. So $154x \equiv c \mod 1001$ is solvable iff $(154,1001)=77 \mid c$.
Therefore $c = 77n, \; n \in \mathbb{N}$.

3. Originally Posted by chiph588@
$ax \equiv b \mod n$ has a solution iff $(a,n) \mid b$. So $154x \equiv c \mod 1001$ is solvable iff $(154,1001)=77 \mid c$.
Therefore $c = 77n, \; n \in \mathbb{N}$.
Since we're working with modulos, I would say that it's 77n mod 1001.

And since 1001/77=13, there are 13 different values for c...