We write $0$ on $7$ vertices of a cube and $1$ on the other vertex . We can select an edge in each step and add $1$ to the numbers written on its ends. Prove gcd of the numbers written on vertices is always $1$.
My "solution" is as follows. Please check it:
Assume for the sake of contradiction that gcd of all the vertices = $d > 1.$ Then after some finite number of moves note that we can always go back to initial configuration. If the gcd of all the vertices is $d>1$ then subtracting multiples of $d$ will result in multiples of $d$, after we subtracted from every vertices, we will be left with multiple of $d$ at the last step. But $d > 1$ so its multiple must also be $> 1$, which contradicts the initial configuration. Thus, gcd must be $1$.
View the vertices of the cube as the set $$V:=\{\pm1\}^3=\{(\pm1,\pm1,\pm1)\},$$ and for each vertex $v\in V$, define its 'weight' $w(v)$ as the product of its coordinates, so that $$w(1,1,1)=1,\qquad w(1,-1,1)=-1,\qquad w(-1,-1,-1)=-1.$$ Denote the number on a vertex $v\in V$ by $n_v$. Then in the initial configuration of the cube you have $n_v=1$ for some $v\in V$, and $n_w=0$ for all other $w\in V$, and taking the weighted sum we have $$w=\sum_{v\in V}w(v)n_v=\pm1.$$ depending on whether the weight of $n_v$. Now note that for any edge of the cube, the two vertices at its ends have opposite weights. This means the weighted sum $$w=\sum_{v\in V}w(v)n_v,$$ remains unchanged when we add $1$ to the vertices at both ends of an edge. It follows that the weighted sum $w$ remains unchanged, so $w=\pm1$. It follows that $\gcd_{v\in V}n_v=1$.