Maximum Face Sum

Since the vertices are numbered from 0 to 7, the maximum value an edge can have is 6+7=13. There are 6 prime numbers less than or equal to 13, and we can list all the options for getting edges to add up to those numbers:

2: 0+2

3: 0+3, 1+2

5: 0+5, 1+4, 2+3

7: 0+7, 1+6, 2+5, 3+4

11: 4+7, 5+6

13: 6+7

There are a total of 13 combinations, and we know there are 12 edges on a cube. So one combination will not be used. Also, each vertex on the cube is part of three different edges. So the numbers 0-7 should each appear in exactly 3 of the sums above. We can see that 0 and 2 appear four times, so the combination 0+2 must be the one that is not used.

Now to arrange the numbers on the vertices. There is only one arrangement (up to rotation and reflection) that uses the pairs in the sums. You can construct the arrangement by placing (for example) 0 at one vertex, and then placing the 3, 5, and 7 at the neighbouring vertices. Then there is a vertex that neighbours the 3 and 5, and (from the list of sums) we see that it must be the 2. You can proceed this way to label the remaining vertices.

Then it is a simple matter to compute the sum of the six faces, and choose the maximal one.

As we can see from the question, sum of any 2 numbers on the edge is prime. Thus, it is better by fixing a number at a vertex. In this case, it is much convenient to fix 7 at a vertex [ I took the largest number 7 because, the probability of finding a prime number decreases with summation of positive numbers]. Now, seeing the numbers, only 4, 0, 6 can find their place beside 7. Fix the numbers 0,4,6 arbitrarily beside 7 now look at common vertex of 4 and 6 other than 7. Only 1 can fit there as the remaining digits [2,3,5] add up to composite numbers. Look at the common vertex to 4 and 0 other than 7, it can take the value 3 alone as the remaining digits [2,5] add up to composite numbers. Look at the common vertex to 1,3 other than 4, it is 2 as sum of the 2 numbers with 5 gives composite numbers. The leftover number is 5, place it in the left over vertex. since the frame of the cube is complete, add up the numbers of each face. the set we get is [18,14,10]. So, obviously the solution is 18

eg: 0+[2,3,5,7] 1+[2,4,6] 2+[3,5,1,0] 3+[2,4,0] 4+[1,3,7] 5+[2,0,6] 6+[1,5,7] 7+[0,4,6] now see each vertices can have only other 3 neighboring vertices..... however we have an additional options for 0 and 2 But since 1,3 and 5 have to combine with 2 so we discard the association between 0 and 2 . So, we have formed the association and drawing the association into the cube we can find out the face with the largest sum is the combination of 1,4,6 and 7 and thus 18

0 cant be situated in such a vertex where 1 or 4 or 6's vertices are directly connected by a side as 0+1, or 0+4 or 0+6 is not a prime,so 0 could share sides with 2,3,5,7 ,but we have to keep in mind that one vertex is connected with exactly 3 sides, so we need optimization by analytical thinking.Likewise 1 must have three friends i.e. sides of vertex labelled by 2,4,6. For 2 there four vertices 0,1,3,5 would be neighbor of him,but we have to decide whose three are our poor 2's neighbor. Similarly for 3,only choices are 0,2,4. Similarly for 4,only choices are 1,3,7 Similarly for 5,only choices are 0,2,6 Similarly for 6,only choices are 1,5,7 Similarly for 7,only choices are 0,4,6 Now apply your imagination Peter,voila! This is now an easy task even for a second grader,as i am a professional engineer,imagination is nothing for me as i know how the potential energy of niagara falls water is used to blend the fruits in my blender

For 7 we have only 3 combinations 7+6 =13 7+0 = 7 7+4 = 11

For 5 we have only 3 combinations 5+2 = 7 5+0 = 5 5+6 = 11

For 6 we have only 3 combinations 6+1 = 7 6+7 = 13 6+5 = 11

Thus putting these in a cube we get the following faces of cubes 6,1,7,4 6,5,1,2 6,7,5,0 3,7,0,4 3,4,2,1 3,2,0,5

Total max being 18

MAXIMUM NO OBTAINED BY ADDING ANY TWO OF THE GIVEN NO'S IS 13, POSSIBLE PRIME NO'S ARE 2,3,5,7,11,13 . CONSIDERING THE VERTICES CONTAINING 7(IT HAS TO CONTRIBUTE TO 3 EDGES),7+0=7,7+4=11,7+6=13. CONSIDERING THE VERTICES CONTAINING 6,6+7=13,6+1=7,6+5=11, CONSIDERING THE VERTICES CONTAINING 4,4+7=11,4+3=7,4+1=5. CONSIDERING THE VERTICES CONTAINING 5,5+6=11,5+2=7,5+0=5. THUS WE GET THE LOCATION OF ALL,THUS BEST PHASE PROVIDING MAXIMUM SUM IS ,7+6+5+0=18 0R7+6+4+1=18

Note that the maximum edge value is 6+7=13. Since there are only three prime numbers between 7 and 13, namely 7,11,and 13, 7 must share an edge with 0,4, and 6. Similarly, there are only three prime numbers between 6 and 13, thus 6 must share an edge with 1 7 and 5. Now consider the number that shares an edge with both 0 and 6 that isn't 7. This must be 5 since 0+1 isn't prime. Thus we have determined the relative positions of 0,5,6,7,1,and 4. The positions of 3 and 2 are fixed because the edges must be prime, and adding up at each face gives 18 as the answer

The first step is to choose a couple numbers and see what numbers it must must be added to in order to obtain a prime number.

Of the 8 numbers, you obtain 3 possible numbers for six of them so you are able to figure out which numbers are on the adjacent vertices through logic.(It isn't necessary to write all of these out. After obtaining a couple of these, you can begin deducing which numbers go where.)

$0: (2,3,5,7), 1: (2,4,5), 2: (0,1,3,5), 3: (0,2,4)$

$4: (1,3,7), 5: (0,2,6), 6: (1,5,7), 7: (0,4,6)$

In my solution, first I found that 4 must have 1, 3, and 7 on the adjacent vertices. I also found that 6 must have 1, 5, and 7 on the adjacent vertices so it must be on the same face as 1, 4, and 7. The remaining two numbers are 0 and 2. 0 cannot go with 1 since it doesn't form a prime number so 2 must go in that vertex leaving 0 to fill the final vertex.

WLOG, choose any vertex to be $1$ . It is connected with $3$ other vertices. They have to be $2,4,6$ in some order. Assign the vertices in any order. Observe that the vertex diagonally opposite to $1$ has to be $0$ [since zero cannot be beside $4,6$ ]. Since it is impossible to link $2$ with zero in the current arrangement, it has to be linked with $3,5,7$ . Trivially check how that can be done. It is also easy to show that this is a unique arrangement other rotation, reflection, etc.

always remember sum of even and odd will you odd(may be prime)but not sum of Odd-odd or even-even-thats the trick;because sum of even -even and sum of odd-odd is always even which defies the value to be Prime. start with 0 and match the other edge vertex with a prime ie..3 ,5,7 then match up with other vertices with the remaining values so that the resulting sum is a prime. never carry even-even or odd odd together.

the possible value are 5,0,6,7 or you may have 1,4,6,7

Step 1. Get the pairs of numbers such that the sum is prime. (0,1),(0,2), (1,2),(1,4),(2,3),(1,6),(2,5),(3,4),(0,7),(5,6),(4,7),(6,7),

The largest sums are 11 and 13. This must be included in the same face. This face has corners 6,7 and 4 with the corner opposite of 7 is 1. The sum of these numbers is 18.

A good way to start this problem is thinking what numbers could stay by side of zero. These numbers just could be the prime numbers 1,3,5,7. After, you have to realize that, between all numbers given, the only one which sumed with the numbers 6,4,2 results in a prime is the number 1. Thinking like this, the problem is resolved.

See the largest sum of a side is 18:

18=6+7+1+4

18+6+7+0+5

We will call a vertex by the number that it is labeled with.

Consider vertex $0$ . The 3 adjoining vertices must be labeled with a prime number, so they must be $2$ , $3$ , $5$ , or $7$ . However, there could be no vertex that is a neighbor of both $2$ and either of $3$ , $5$ or $7$ , since one of the sums would be an even number greater than $2$ , thus not prime. Hence $2$ is not adjoining next to $0$ . So the vertices adjoining $0$ are $3$ , $5$ and $7$ .

The only vertex that can be adjoining both $3$ and $7$ is $4$ . The only vertex that can be adjoining both $3$ and $5$ is $2$ . The only vertex that can be adjoining both $5$ and $7$ is $6$ . The remaining vertex is $1$ , which adjoins $2$ , $4$ and $6$ , and we can check that the sums are prime.

Therefore the way to label the cube is unique (up to permutations of the cube).

It is easy to check that the maximum sum of the 4 numbers on a face is $7 + 6 + 5 + 0 = 7 + 6 + 4 + 1 = 18$ .

The maximum of surface value is 18. There will be two such surfaces.

I drew a cube on paper and labelled the vertices in such a manner so that the sum of any two numbers on the same edge is a prime.

the face which has 7,2,4 and 5 as vertices has sum maximum.