Prove that if $12\mid{a}$ then $4\nmid{b}$ or $4\mid{(8+2a+5b)}$.
I have gotten thus far:
Since $12\mid{a}$, then $a=12k$ for some integer $k$. Substitute $a=12k$ into $(8+2a+5b)$ to get $8+24k+5b$.
Not sure what to do from here, any help is appreciated, thanks!
Assume $12|a$ and $4|b$ , otherwise nothing has to be proven, then , we can write $a=4s$ and $b=4t$ with intgegers $s,t$
Then, we have $$8+2a+5b=8+8s+20t=4(2+2s+5t)$$ hence $4$ divides $8+2a+5b$