If $n$ is an odd integer, which is equivalent? Given $n$ is an odd integer, which of the following statements is equivalent?
 I narrowed it down to two, but they are both true 


*

*$7n$ is odd

*$6n$ is even


Since $n$ is odd there exists an integer $k$ such that $n = 2k+1$. Plugging in $2k+1$ into the first statement we get 
$$7n = 7(2k+1) = 14k + 7 = (14k + 6) + 1 = 2(7k + 3) + 1$$
which is odd. 
$-6n$ already has a factor of two so it's even. So I not sure which one to pick.
What do you do in this situation, when more than one answer seems correct? 
 A: They're not  both equivalent to ‘$n$ is odd’: 
The first is truly equivalent, since you can prove conversely that if $7n$ is odd, then $n$ is odd.
The second is implied by ‘$n$ is odd’ – actually, it is also true also if $n$ is even. So $-6n$ is even  does not imply whatever.
A: I think there is some confusion between "$p$ implies $q$" and "$p$ is equivalent to $q$".
"$p$ implies $q$" means one can derive $q$ if one assumes $p$. In your example, you are taking $p$ to be "$n$ is odd". From this hypothesis, it is indeed true that both $7n$ is odd and $6n$ is even.
"$p$ is equivalent to $q$" is a sort of shorthand for "$p$ implies $q$ and $q$ implies $p$". We already dealt with the first half, but what about the second? 
If we assume $6n$ is even, can we conclude $n$ is odd? We cannot; $6n$ is always even for any integer $n$ (including even ones).
If we assume $7n$ is odd, can we conclude $n$ is odd? Yes, but I will leave the details to you. (It will look much like the proof you already gave.)
Taking all this together, "$n$ is odd" implies "$6n$ is even" but they are not equivalent, while "$n$ is odd" is equivalent to "$7n$ is odd".
