2007 AMC 10A Problem 11

Below is the professionally curated solution for Problem 11 of the 2007 AMC 10A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2007 AMC 10A solutions, or check the answer key.

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Concepts:double counting cube geometry

Difficulty rating: 1280

11.

The numbers from $1$ to $8$ are placed at the vertices of a cube in such a manner that the sum of the four numbers on each face is the same. What is this common sum?

$14$

$16$

$18$

$20$

$24$

Solution:

Each vertex belongs to exactly three faces, so summing the numbers over all six faces gives $3(1 + 2 + \cdots + 8) = 3 \cdot 36 = 108.$

There are six faces, so the common sum is $108 \div 6 = 18.$

Thus, the correct answer is C.

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