Given that k is a positive integer less than 6, how many values can k take on such that 3k\(\equiv\)k (mod6)has no solutions in x?
Given that k is a positive integer less than 6, how many values can k take on such that 3k=k (mod6)has no solutions in x?
\(\frac{3k}{6}=a+\frac{n}{6}\quad and \quad \frac{k}{6}=b+\frac{n}{6}\quad \\\text{where a, b and n are pos integers and }k<6\)
if k=1 it doesn't work
if k=2 it doesn't work
if k=3 it works
if k= 4 it doesn't work
if k= 5 it doesn't work
so k=3
k can only take on 1 value and that value is 3