Source : Burton, Elementary number theory (7th edition)
Show that any integer of the form $6k + 5$ is also of the form $3 j + 2$, but not conversely.
As to the $(\implies)$ part :
$n= 6k+5$
$\implies n = 3(2k')+5 \space $ for some $ k' \in \mathbb Z$
$\implies n = 3(2k')+3+2 $
$\implies n = 3(2k'+1)+2 $
$ \implies n = 3j +2 \space $ with $ j = 2 k'+1 $
As to the " not conversely part".
It seems to suffice to find a number of the form $3j+2$ that does not leave a remainder of $5$ when divided by $6$.
Consider number $20 = 3(j)+2 $ where $j=6$.
We have also $ 20 = 6(3) +2$ .
My question is : is the counterexample method adequate and, if so, did I apply it correctly? Is stating that $20$ can be written with a remainder of $2$ when divided by $6$ sufficient to show that it cannot be written with a remainder of $5$.