If j, k, and n are consecutive integers such that $0<j<k<n$ and the units (ones) digit of the product jn is 9, what is the units digit of k ? 

There are only a few ways you can make a product have a units digit of $9$. Either both numbers you’re multiplying have to be $3$, or one has to be $1$ and one has to be $9$. Since we’re dealing with consecutive integers, $j$ and $n$ can’t both end in $3$, so they’re going to have to end in $1$ and $9$. 

It’s important to remember that although we only really care about the units digits in this problem, we’re dealing with numbers that might (or in fact, must) be $2$ or more digits. That’s why you can have $k^{\prime }s$ units digit be $0$. 

Say $j=19$, $k=20$, and $n=21$. Then $jn=(19)(21)=399$ (units digit is $9$), and the units digit of $k=0$.

Hence option A is correct.