A solid cube of side length $1$ is removed from four corners of a solid cube of side length $4$, no two of the four corners adjacent to each other. How many edges does the remaining solid have? {\small The solid after the four corner cubes are removed.}

1. The $8$ vertices of a cube split into two sets of $4$ by coordinate-parity, with no two vertices in the same set adjacent. Removing the $4$ corners of one such set: each removal deletes $1$ old vertex and adds $7$ new ones, giving $V=8-4+4(7)=32$.
2. Since adjacent cube-vertices always have opposite parity, every one of the $12$ original edges touches exactly one removed corner, so all $12$ survive (shortened at one end). Each corner-removal again contributes $9$ new edges (the removals don't interact, since same-parity corners aren't adjacent), giving $12+4(9)=48$.
3. (Check via Euler's formula: $F=2-V+E=2-32+48=18$, matching the $6$ original faces (each now an octagon, with two opposite corners notched) $+4\times3$ new square faces $=18$.) So the answer is $48$.