A probability problem by Lakshika Rathi

Visualising the $3$ by $3$ cube, notice that there are

• $8$ cubes with $3$ exposed sides
• $12$ cubes with $2$ exposed sides
• $6$ cubes with $1$ exposed sides
• $1$ cube with no exposed side (the cube in the core of the big cube)

You can verify the above by adding $8+12+6+1$ to obtain $27$ cubes altogether.

Notice that all sides above are non-opposite (which means that they do not follow the rule that opposite sides add up to $7$ .

Thus smallest sum of all numbers on the outside faces $8(1+2+3) + 12(1+2) + 6(1) = 90$ .

There are altogether $6\times 9=54$ faces on the large cube.

• The smallest number for the $6$ centre die face is $1$ , therefore, $N_1=6\times 1 = 6$

• The smallest two numbers for the two faces of the dice in the middle of the $12$ edges are $1$ and $2$ , therefore, $N_2=12\times (1+2)=36$

• The smallest three numbers for the three faces of the dice at the $8$ corners are $1$ , $2$ and $3$ , therefore, $N_3=8\times (1+2+3)=48$

• Therefore, the smallest possible sum, $S_{min}=N_1+N_2+N_3=6+36+48=\boxed{90}$

In such a configuration, the cubes having at least one side are 26, out of which 6 are centre pieces with one face each, 12 edge pieces with 2 faces each and 8 corner pieces with 3 faces each. The minimum sum of faces possible to attain for centre, edge and corner pieces are $1$ , $(1+2)$ and $(1+2+3)$ respectively.Hence, the minimum sum of all the faces of the cube so formed are $(1*6)+(3*12)+(6*8) = 90$ .

This is like a rubiks cube. You have 8 dices at 8 vertices that reveal 3 faces, least total for these dices is 6 (3+2+1). The centre die in each face reveals only 1 face hence 6 sides of the cube 6 centre die. All other dices reveal 2 faces the least total for then is 3 (2+1).

Total is 8x6+6x1+12x3=90