Maths Olympiad Prep

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Problem 847

AMC 12 late, AIME early
Combinatorics Difficulty 4.6 Multiple choice Gara di Febbraio · Italy

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, 1414, 2222, 3030, 3838, 4646, 5454. What is the sum of all the faces of the dice which, having been glued together, are no longer readable?

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Official solution

Solution:

The answer is (E)(E). The sum of the values on the faces of a die is 1+2+3+4+5+6=211+2+3+4+5+6=21. The sum over all faces, visible and not, is therefore 2721=56727 \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)=567204=363567-(14+22+30+38+46+54)=567-204=363.

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