How many positive integers n satisfy 127 = 3 (mod n)? n = 1 is allowed.
n = 4, 31, 62, 124
Subracting \(3\) from both sides gives \(124=0 \pmod n\). This means that \(n\) is a factor of \(124\). Since \(124=2^2 \cdot 31^1\), the amount of factors \(124\) has is \((2+1)(1+1)=6\), which should be the answer.
+118 
