On a standard 6 sided dice, the numbers on opposite faces add up to 7. Sue has 27 standard dice, which she stacks to form a large 3x3x3 cube. She then adds all the numbers of the outside faces of her large cube.
What is the smallest possible sum that Sue can get?
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There are altogether 6 × 9 = 5 4 faces on the large cube.
The smallest number for the 6 centre die face is 1 , therefore, N 1 = 6 × 1 = 6
The smallest two numbers for the two faces of the dice in the middle of the 1 2 edges are 1 and 2 , therefore, N 2 = 1 2 × ( 1 + 2 ) = 3 6
The smallest three numbers for the three faces of the dice at the 8 corners are 1 , 2 and 3 , therefore, N 3 = 8 × ( 1 + 2 + 3 ) = 4 8
Therefore, the smallest possible sum, S m i n = N 1 + N 2 + N 3 = 6 + 3 6 + 4 8 = 9 0
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 ∗ 1 2 ) + ( 6 ∗ 8 ) = 9 0 .
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
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Visualising the 3 by 3 cube, notice that there are
You can verify the above by adding 8 + 1 2 + 6 + 1 to obtain 2 7 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 ) + 1 2 ( 1 + 2 ) + 6 ( 1 ) = 9 0 .