In the following equation, each of the letters represents uniquely a different digit in base ten: 
\[(YE) \cdot (ME) = TTT\]
The sum $E+M+T+Y$ equals 
$\textbf{(A)}\ 19 \qquad \textbf{(B)}\ 20 \qquad \textbf{(C)}\ 21 \qquad \textbf{(D)}\ 22 \qquad \textbf{(E)}\ 24$
The right side of the equation can be rewritten as $111T = 37 \cdot 3T$.  With trial and error and prime factorization as a guide, we can test different digits of $T$ to see if we can find two two-digit numbers that have the same units digit and multiply to $111T$.
The only possibility that works is $37 \cdot 27 = 999$.  That means $E+M+T+Y = \boxed{\textbf{(C)}\ 21}$.