For every integer a, b, c, if 3c is not divisible by a, then b is not divisible by a or 3c is not divisible by b I can't seem to figure this proof out. How would I prove this by contradiction and contraposition?
I tried doing it by saying $3c=bk=(ak)k=a(kk)$
since $b=ak$, $3c=bk$ for some interger $k$
$3c=a(kk)$ contradicts "3c is not divisible by a"
contraposition would follow suit since we prove that $3c=a(kk)$
What am I doing wrong?
 A: First figure out how to say this by contrapositive.
The statement is of the form  $\lnot P \to \lnot Q \lor \lnot R$.
Now $\lnot Q \lor \lnot R \iff \lnot (Q \land R)$ so the statment that you are trying to prove is equivalent to
$\lnot P \to \lnot(Q \land R)$ and that is equivalent to the contrapositive:
$(Q\land R) \to P$.
So it is equivalent to prove the following:

If $b$ is divisible by $a$ and $3c$ is divisible by $b$ then $3c$ is divisible by $a$.

That is almost too easy to deserve discussion:
Pf:  $a|b$ so there exist an integer $k$ so that $b = ka$.  And $b|3c$ so there exists an integer $j$ so that $3c = jb$.  So $3c =j(ka) = (jk)a$.  $jk$ is an integer so $a|3c$.  QED.
To show that is equivalent to the title statement:
Pf:  Note:  If $a|b$ and $b|3c$ then we would have $a|3c$.  But we were given that $a \not \mid 3c$. So we can't have both $a|b$ and $b|3c$.  So we must have either $a\not \mid b$ or $b\not \mid 3c$.
That's it.  We're done.  Let's go home and eat lunch.
A: If b is divisible by a and 3c is divisible by b then 3c is divisible by a, a contradiction . Thus one of them must be true.
A: Hint : The contraposition of this proposition is "if $b$ is dividible by $a$ and $3c$ is dividible by $b$, then $3c$ is dividible by $a$".
