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.