Let $n$ be an integer. If the tens digit of $n^2$ is 7, what is the units digit of $n^2$? 
Let $n$ be an integer. If the tens digit of $n^2$ is 7, what is the units digit of $n^2$?

So $n^2 \equiv 7 \pmod{100}$? If this is the case then this can be written as $n^2 = 100k +7$, where $k \in \Bbb Z.$
Here one can see that no matter what the choice of $k$, the units digit will be $7$. Thus $n^2 \equiv 7 \pmod{10}.$ However this was wrong. The correct answer is $\textbf{6}.$
What am I doing wrong here? It seems that $n^2 \equiv 7 \pmod{100}$ doesn't hold. If the tens digit is $7$ should I have that $n^2 \equiv 7k \pmod{100}$, where $k$ represents the unit digit of $70$ and not a multiplication?

Hint You are looking for adigit $k$ such that
$$n^2 \equiv 70+ k \pmod{100}$$
By the Chinese Remainder Theorem this is equivalent to
$$n^2 \equiv 2+ k \pmod{4}\\
n^2 \equiv  k-5\pmod{25}
$$
The quadratic residues modulo $4$ are $0,1$, therefore $k \in \{ 2,3, 6,7 \}$. You have now to figure for which of those $k-5$ is a quadratic residue modulo $25$.
