Each vertex of a cube is to be labelled with an integer 1 through 8, without repetition, such that sum of numbers of the four vertices of a face is the same for each face. Arrangements that can be obtained through rotations of the cube are considered to be the same. How many different arrangements are possible?

My attempt: Fix 8 on E.

enter image description here

Then, D+C=F+G (Since H is common to both faces)

Similarly, C+H=A+F and D+A=H+G

We can extend this analogy to other sides, and we see that D+E=B+G, E+H=A+B, E+F=B+C.

This means that {1,2}, {1,3}, {7,8}, {8,6} cannot be on one line segment since there are no other distinct numbers which add up to give 3,4,15 or 14 respectively.

How can I get the other conditions?

Edit: As @John and @Alexander mentioned in the comments, the sum of numbers on each face should be 18. The possible sets of numbers on opposite faces will be {8,7,1,2}, {3,4,5,6}; {8,6,1,3}, {4,5,7,2}; {8,5,2,3}, {1,4,7,6} and {8,5,1,4}, {2,3,6,7}.

Case 2 will not have any possibilities since there are no 2 numbers except for 8,3 which add up to 11. In case 4, 4 must be opposite to 8 and in case 1, 7 must be opposite to 8.

Now we can try to count it on a case by case basis, but I can't think of a foolproof method without repetitions. Can someone help :)

  • 5
    $\begingroup$ Let $S$ be the sum of vertex values common to all 6 faces. Since each vertex participates in 3 faces, the "sum of faces" $6S$ must be $3\times(1+2+3+4+5+6+7+8) = 108$. Hence the sum on each face can only be $S = 108/6 = 18$. $\endgroup$ Sep 24, 2021 at 4:26
  • $\begingroup$ @JohnBarber This means that the possible sets of vertices on the faces containing $8$ must be one of $\{8,7,2,1\}$, $\{8,6,3,1\}$, $\{8,5,4,1\}$, $\{8,5,3,2\}$. The opposite face will then be one of $\{6,5,4,3\}$, $\{7,5,4,2\}$, $\{7,6,3,2\}$, $\{7,6,4,1\}$. $\endgroup$ Sep 24, 2021 at 4:44
  • 1
    $\begingroup$ @John You could use just two faces. $\endgroup$
    – Moti
    Sep 24, 2021 at 5:02
  • $\begingroup$ So, if $7$ is not on any face with $8$, then $8$ should be on 2 faces with $5$, so $5$ should be adjacent to $8$. That yields 1 labeling up to symmetries. Alternatively, $7$ is on some face with $8$. Then, as OP noted, it cannot be adjacent to $8$, so it's diagonally opposite on that face. The other two vertices, then, are $1$ and $2$. So, $7$ is adjacent to $1$. Therefore, the other 2 vertices on the 71 face are $6$ and $4$. That yields 2 labelings, since the $64$ edge can be oriented in two different ways. I think that's it, up to rotations and reflections. Now get reflections out of it. $\endgroup$ Sep 24, 2021 at 5:06

2 Answers 2


Note that the sum of the numbers on each face must be $18$ because $\frac{1+2+\cdots+8}{2}=18$.

So now consider the opposite edges (two edges that are parallel but not on the same face of the cube); they must have the same sum value too. Now think about the points $1$ and $8$. If they are not on the same edge, they must be endpoints of opposite edges, and we should have $1+X=8+Y$. However, this scenario would yield no solution for $[2,7]$, which is a contradiction. The points $1$ and $8$ are therefore on the same side and all edges parallel must also sum to $9 .$ Now we have $4$ parallel sides $1-8,2-7,3-6,4-5$. Thinking about $4$ endpoints, we realize they need to sum to 18. It is easy to notice only $1-7-6-4$ and $8-2-3-5$ would work. So if we fix one direction $1-8($ or $8-1)$ all other 3 parallel sides must lay in one particular direction. $(1-8,7-2,6-3,4-5)$ or $(8-1,2-7,3-6,5-4)$ Now, the problem is the same as arranging 4 points in a two-dimensional square, which is $\frac{4 !}{4}=6$.

It is also available here. There is also a video solution here.


The answer is $6$.

Identify the set of vertices with $\{(a,b,c)\mid a,b,c\in \{0,1\}\}$. I write $C(a,b,c)$ for the color of the vertex $(a,b,c)$. Suppose that $C(0,0,0)=1$ and $C(1,1,1)=1+x$. Then $$\sum_{i,j}C(i,j,0)=\sum_{k,l}C(k,1,l)$$ yields $$C(1,0,0)=C(0,1,1)+x.$$ Similarly, we get $$C(0,1,0)=C(1,0,1)+x$$ and $$C(0,0,1)=C(1,1,0)+x.$$ In summary, $C(a,b,c)=C(1-a,1-b,1-c)+x$ holds for all $(a,b,c)$ with $a+b+c$ odd. Conversely, any labeling with this property satisfies the required condition. There are three cases:

  • $x=1$ and $\{C(0,1,1), C(1,0,1), C(1,1,0)\}=\{3,5,7\}$.
  • $x=2$ and $\{C(0,1,1), C(1,0,1), C(1,1,0)\}=\{2,5,6\}$.
  • $x=4$ and $\{C(0,1,1), C(1,0,1), C(1,1,0)\}=\{2,3,4\}$.

For each cases we have two distinct labelings, so there are exactly $3×2=6$ labelings.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .