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?

MobiusLoops
Sep 14, 2018

#2**+2 **

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

Melody
Sep 15, 2018