What is the tens digit of $7^{2011}$?
$\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad\textbf{(E) }7$
Since we want the tens digit, we can find the last two digits of $7^{2011}$.  We can do this by using modular arithmetic.
\[7^1\equiv 07 \pmod{100}.\]
\[7^2\equiv 49 \pmod{100}.\]
\[7^3\equiv 43 \pmod{100}.\]
\[7^4\equiv 01 \pmod{100}.\]
We can write $7^{2011}$ as $(7^4)^{502}\times 7^3$.  Using this, we can say:
\[7^{2011}\equiv (7^4)^{502}\times 7^3\equiv 7^3\equiv 343\equiv 43\pmod{100}.\]
From the above, we can conclude that the last two digits of $7^{2011}$ are 43.  Since they have asked us to find the tens digit, our answer is $\boxed{\textbf{(D)}\ 4}$.
-Ilovefruits