3. The six edges of the tetrahedron $ABCD$ are 7, 13, 18, 27, 36, 41, and $AB=41$. Then $CD=(\quad)$.
(A) 7
(B) 13
(C) 18
(D) 27
3. B.

In a tetrahedron, except for $CD$, all other edges are adjacent to $AB$. If the edge of length 13 is adjacent to $AB$, let's assume $BC=13$. According to the triangle inequality, we have
$$
\begin{array}{l}
AC \notin\{7,18,27\} \Rightarrow AC=36 \Rightarrow BD=7 \\
\Rightarrow\{AD, CD\}=\{18,27\} .
\end{array}
$$

Thus, in $\triangle ABD$, the sum of two sides is less than the third side, which is a contradiction.
Therefore, only $CD=13$.
On the other hand, a tetrahedron $ABCD$ with $AB=41, CD=13$ can be constructed (for example, by taking $BC=7, AC=36, BD=18, AD=27$).