(3) The six edges of tetrahedron $ABCD$ have lengths $7, 13, 18, 27, 36, 41$, and $AB=41$. Then $CD$ is ( ).
(A) 7
(B) 13
(C) 18
(D) 27
(3) In the 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, $AC \notin \{7, 18, 27\}$, so $AC=36, BD=7$, which gives $\{AD, CD\}=\{18,27\}$. 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 $BC=7, AC=36, BD=18, AD=27$. Hence, the correct choice is B.