really? the proof is pretty straightforward though. anyway it's a challenge so... keep trying!!

Why don't you post your prove Lucky? It may be inelegant, but it's better than nothing!

I have a solution that, I believe, is the best for this problem: the lowest value of a edge is 3 and the higher is 15. in the sum of all edges we have 3 times each edge, so the sum of alledges is ( \frac{(1+8)8}{2} )3 = 108 now the higher 12 values between 3 and 15 are the number from 4 to 15 wich sum is 104.

Bajoukx wrote: now the higher 12 values between 3 and 15 are the number from 4 to 15 wich sum is 104. No. Their sum is 114 Try again!

such a stupid mystake... well, I try again... between 3 and 15 we have 13 numbers, we need 12 diferent of them to the values of the sum in the edges. and we need that the sum of these 12values = 108, so these 12 numbers are the numbers between 3 and 15 less the 9. the only way to do 15= 8+7 14= 8+6 13= 8+5 since 6 and 7 can be in the the corners of the same edge and now we see that we can't do the 12 since we just can use each number three times and we can't put the 5 and 7 in the corners of the same edge... I hope now it's everything correct

perfect!

I think it is still not perfect. You should prove that the 12 values include 12,13,14,15. (It is easy though.)

Johan Gunardi wrote: I think it is still not perfect. You should prove that the 12 values include 12,13,14,15. (It is easy though.) Bajoukx wrote: such a stupid mystake... well, I try again... between 3 and 15 we have 13 numbers, we need 12 diferent of them to the values of the sum in the edges. and we need that the sum of these 12values = 108, so these 12 numbers are the numbers between 3 and 15 less the 9. the only way to do 15= 8+7 14= 8+6 13= 8+5 since 6 and 7 can be in the the corners of the same edge and now we see that we can't do the 12 since we just can use each number three times and we can't put the 5 and 7 in the corners of the same edge... I hope now it's everything correct he does prove it

No he didn't. Between 3 and 15, inclusive, there are 13 numbers. We only choose 12 of them. We have to prove that 12, 13, 14 and 15 are chosen.

What he said was that the sum of the 12 numbers must be 108. We start out with the 13 numbers: 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15. (Because each corner is labeled from 1-8, and no numbers are repeated, the minimum value for an edge is 3 (1+2), and the maximum, 15 (8+7).) However, the sum of these 13 numbers equals 117, or 108 + 9. Therefore, the number 9 must be excluded. (Apart from grammar mistakes, all of this was stated.) If you have any qualms, please ask them.

Oh, I see. I misunderstood him. He said that the numbers are 'less than 9'. So I thought he was referring to the values of the vertices.

I don't blame you. I had to reread that a few times to figure out what he meant, too...

We prove by contradiction. Suppose it is possible. Note that the minimum value of the sum of vertices on an edge is $1+2=3$, and the maximum is $7+8=15$. Hence there are $15-3+1=13$ possible values of the sum on an edge. Since there are $12$ edges on the cube, there is exactly one possible value of the sum on an edge which is not used. Observe that the sum of all edges is $3(1+2+\ldots+8)=108$ and the sum of all possible values is $3+4+\ldots+15=117$. Hence, there is no edge with value $117-108=9$. Therefore there exist edges with values $15,14,13,12$. Clearly, the edge with sum $15$ must have $7$ and $8$ on its vertices. Similarly, $14=6+8$. Therefore, $6$ and $7$ do not share an edge. Hence $13$ must be formed with $5$ and $8$. Now $12=4+8$ or $5+7$. Since $8$ already has $3$ edges connected to it, $12=4+8$ is immediately rejected. Also, since $5$ and $7$ do not share an edge, $12$ cannot be formed. Hence we have a contradiction, and it is impossible.

Among $1-8$, there are $13$ different sums that can be formed, all these sums give a total sum of $117$ and since each vertex is counted trice, we need a total sum of $108$ which means $9$ must be excluded from these sums. Now, we know $(1,2), (1,2), (1,4)$ must share an edge but then the sum of $6$ can't be formed. Thus, it is impossible.

mathmax12's solution (which he told me to post): The sum of the numbers can be anywhere from $3-15$, adding the possible sums up we get $117$ so we need to exclued $9$, so it is not possible.