3. For the tetrahedron $ABCD$, the 6 edge lengths are 7, 13, 18, 27, 36, 41, and it is known that $AB=41$, then $CD=$ $\qquad$
3. 13 .

Solution: 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, we know $AC \notin \{7,18,27\}$, thus $AC=36$, $\Rightarrow BD=7$, $\Rightarrow \{AD, CD\}=\{18,27\}$, then 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 $BC=7, AC=36, BD=18, AD=27$.