Could the sum of all $26$ numbers be equal to $1234$ 
Suppose we write a natural number at each vertex of a cube. Then at the midpoints of each edge we write the sum of the two numbers that are at the ends of the edges. Finally, in the middle of each face we write the sum of he four numbers that are at the vertices of this face. Could the sum of all $26$ numbers be equal to $1234$? 

I know that I will have $6$ face numbers $12$ edge numbers and $8$ vertex numbers. I used $a,b,c,d,e,f,g,h$ as my vertex points then I tried to see what would happen when I add them all up so I got if I didn't make a mistake was $7a+8b+8e+7c+5d+8f+7g+6h$ now I'm stuck and unsure if this was the right approach. 
 A: There's a lot of repetion  involved.
The 8 vertices are $a_i$ so the some of the vertices are $V= \sum_{i=1}^8 a_i$.  Each of the $12$ midpoint  is the sum of two vertices and each vertex is in three midpoints.  So the the sum of the midpoints is $M = 3\sum a_i = 3V$.
The middle of each of the $6$ face is contains the four vertices and each vertex is in $3$ faces so the sum of the vertices included in the faces is $F = 3V$. 
So the total of all the numbers is $V + M + F = V + 3V + 3V = 7V$.
Does $7$ divide $1234$?  It does not.  So such a sum is impossible.  Buy any multiple of $13$ that is larger or equal to  $7*8 = 56$ is possible.
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Oops.  I first thought the number in the middle of each face was the sum of the vertices and the sum of the midpoints.  That would make $F_v = 3V$ and $F_m = 2M$ (because each midpoint is in two faces).  So $F = 3V + 2*3V = 9V$ and $V+M+F = V + 3V + 9V=13$. 
As such it is still impossible as $13$ does not divide $1234$. However we can get this number to equal $1235$ by letting vertices $a_1... a_7 = 1$ and letting $a_8 = 88$. The sum of the vertices will be $95$. Then the midpoints of all but three of the edges would be $2$ and three of the edges would be $89$. 
So the sum of the edges would be $9*2 + 3*89 = 285$.
Three of the faces would be have vertices $1,1,1,1$ and edges $2,2,2,2$ for a value of $ 12$.  The other three faces would have vertices of $1,1,1,88$ and edges of $2,2,89,89$ for a value of $91 + 2+2+89+89 = 273$. 
So the sum of the faces would be $3*12 + 3*273 = 855$.
So the total would be $95 + 285 + 855 = 1235$
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Alternative answer.  Suppose the sum of the numbers is $K$.  Then change one vertex one higher.  The will add $1$ to the one vertex.  It will add $1$ to each of the three lines containing this vertix.  And it will add $1$ to each of the three faces that meet at this vertix.  That will increase the total by $7$ to $K + 7$. 
Likewise decreasing exactly on vertex by $1$ will result is $K -7$.
So via induction the result must be $M + 7k$ where $M$ is how much the sum will be if every vertex is $1$, and $k$ is the total number of vertices modified and by how much.
If each vertex is $1$ then each edge is $2$ and each face is $4$ so $M = 8 + 12*2 + 6*4 = 56$.  
So the only sums are $56 + 7k = 7(k + 8)$ for any natural number $k$.
A: If a vertex has a value v1, it will appear 7 times in the equation. 1 for the vertex, 3 associated edges, and 3 faces. With symmetry you have 7 times the sum of all vertices. As the number 1234 is not divisible by 7, this is not possible.
A: Taking vertex points as $a,b,c,d,e,f,g,h$, and adding them according to the problem leads to the equation 
$7(a+b+c+d+e+f+g+h)$
$i.e.$ every variable adds exactly 7 times.clearly this addition cannot gives the result $1234$, as $1234$ is not divisible by 7.
Hence the sum cant be equal to 1234, irrespective of any natural values of variables $a,b,c,d,e,f,g,h$.
