The faces of a cube contain the number 1, 2, 3, 4, 5, 6 such that the sum of the numbers on each pair of opposite faces is 7. For each of the cube’s eight corners, we multiply the three numbers on the faces incident to that corner, and write down its value. (In the diagram, the value of the indicated corner is 1 x 2 x 3 = 6.) What is the sum of the eight values assigned to the cube’s corners?
Problem
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Tags: geometry, 3D geometry
talkinaway
05.11.2012 01:14
We count the $4$ vertices adjacent to the $1$. Place the $1$ on top. The numbers on the side of the die, going in one direction, are $2, 3, 5, 4$. We know this because: A) $6$ must be on the bottom so it's not there, and B) Of the remaining numbers $2, 3, 4, 5$, the pairs are $2 + 5 = 3 + 4 = 7$, so they can't touch each other.
So, the sum of our four products on top is $1 \cdot (2 \cdot 3 + 3 \cdot 5 + 5 \cdot 4 + 4 \cdot 2)$
Now we look at the bottom face of the die, a $6$. We get the same order of adjacent faces: $2, 3, 5, 4$. We go counterclockwise with $1$, and clockwise with $6$ (or vice versa - I don't have a die with me), but the direction you go is irrelevant.
So, our four products on bottom are $6 \cdot (2 \cdot 3 + 3 \cdot 5 + 5 \cdot 4 + 4 \cdot 2)$
Adding those gives $7 \cdot (2 \cdot 3 + 3 \cdot 5 + 5 \cdot 4 + 4 \cdot 2)$, which simplifies to $7 \cdot 49$, or $\boxed{343}$.