I am having trouble understanding the solution from IMO Compendium. Aren't we trying to look for hamiltonian cycles rather than eulerian ones? Thanks Solution by IMO Compendium: Let us set a coordinate system denoting the vertices of the block. The vertices of the unit cubes of the block can be described as {(x, y, z) | 0 ≤ x ≤ p, 0 ≤ y ≤ q, 0 ≤ z ≤ r}, and we restrict our attention to only these points. Suppose the point A is fixed at (a, b, c). Then for every other necklace point (x, y, z) numbers x − a, y − b, and z − c must be of equal parity. Conversely, every point (x, y, z) such that x − a, y − b, and z − c are of the same parity has to be a necklace point. Consider the graph G whose vertices are all such points and edges are all diagonals of the unit cubes through these points. In part (a) we are looking for an open or closed Euler path, while in part (b) we are looking for a closed Euler path. Necklace points in the interior of the (p, q, r) box have degree 8, points on the surface have degree 4, points on the edge have degree 2, and points on the corner have degree 1. A closed Euler path can be formed if and only if all vertices are of an even degree, while an open Euler path can be formed if and only if exactly two vertices have an odd degree. Hence the problem in part (a) amounts to being able to choose a point A such that 0 or 2 corner vertices are necklace vertices, whereas in part (b) no corner points can be necklace vertices. We distinguish two cases. (i) At least two of p, q, r, say p, q, are even. We can choose a = 1, b = c = 0. In this case none of the corners is a necklace point. Hence a closed Euler path exists. (ii) At most one of p, q, r is even. However one chooses A, exactly two necklace points are at the corners. Hence, an open Euler path exists, but it is impossible to form a closed path. Hence, in part (a), a box can be made of all (p, q, r) and in part (b) only those (p, q, r) where at least two of the numbers are even.

Solution from Twitch Solves ISL: For (a), the answer is always yes; whereas for (b), the answer is only if at least two of $p$, $q$, $r$ are even. Let the box $\mathcal B$ be located in space spanning $(0,0,0)$ to $(p,q,r)$. Then the string can move from $(x,y,z)$ to $(x \pm 1, y \pm 1, z \pm 1)$. So, the string is going to visit only points of two parity classes in ${\mathbb F}_2 \times {\mathbb F}_2 \times {\mathbb F}_2$, say $(\varepsilon_x \bmod 2, \varepsilon_y \bmod 2, \varepsilon_z \bmod 2)$ and $(1+\varepsilon_x \bmod 2, 1+\varepsilon_y \bmod 2, 1+\varepsilon_z \bmod 2)$. However, every unit cube has a string through it, so we find that the string visits every point of this form in $\mathcal B$. Let us denote the set of points by $V$ (which depends on the choice of $(\varepsilon_x, \varepsilon_y, \varepsilon_z)$). Hence the question is asking when we can have a Eulerian path or circuit on the resulting graph $G$ (whose vertex set is $V$ and whose edges are the string). However, the degrees of vertices of $G$ are always even except for the corners of $\mathcal B$, i.e.\ the eight points \[ \Big\{ (0,0,0), \; (p,0,0), \; (0,q,0), \; (0,0,r), \; (p,q,0), \; (p,0,r), \; (0,q,r), \; (p,q,r) \Big\}. \] We can always ensure there are at most two corners in $V$ by choosing $\varepsilon_x$, $\varepsilon_y$, $\varepsilon_z$ randomly; the expected number of corners is $\frac14 \cdot 8 = 2$. We can ensure there are zero corners if at least two of $(p,q,r)$ are even; otherwise we cannot. Hence by the classical theorem on Eulerian paths and circuits, (a) is possible always, (b) is possible when at least two of $(p,q,r)$ are even.