Competition problem — Combinatorics

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Combinatorics Difficulty 4.6 AIME Find the answer Italy

Problem:

Andrea glues 27 ordinary 6-faced dice together to form a large cube. The dice are oriented so that the sums of the values readable on each face of the cube are, in some order, $14$ , $22$ , $30$ , $38$ , $46$ , $54$ . What is the sum of all the faces of the dice which, having been glued together, are no longer readable?

Pick one

Solution

Solution:

The answer is $(E)$ . The sum of the values on the faces of a die is $1+2+3+4+5+6=21$ . The sum over all faces, visible and not, is therefore $27 \cdot 21=567$ . To obtain the sum on the hidden faces we can subtract from this number the sum of the numbers written on the visible ones: the answer is therefore $567-(14+22+30+38+46+54)=567-204=363$ .

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Source: MathNet, licensed CC-BY-4.0. Statement translated into English from it; metadata (topic, difficulty) added by this project.