+0

# modular arithmetic

0
45
2

How many positive integers n satisfy 127 = 3 (mod n)? n = 1 is allowed.

Nov 26, 2020

#1
0

n = 4,  31,  62,  124

Nov 26, 2020
#2
+105
0

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.

Nov 26, 2020