Hey RB!
If \(127 \equiv 7 \pmod{n}\), then n is a divisor of 127 - 7 = 120.
The prime factorization of 120 is \(2^3 \cdot 3 \cdot 5\)
which has \((3 + 1)(1 + 1)(1 + 1) = 16\) positive divisors.
Therefore, there are 16 possible values of n.
I hope this helps,
Gavin.
+235 
