2023 AMC 10A Problem 6

Below is the professionally curated solution for Problem 6 of the 2023 AMC 10A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2023 AMC 10A solutions, or check the answer key.

All of the real AMC 8, AMC 10, AMC 12, and AIME problems in our complete solution collection are used with official legal permission of the Mathematical Association of America (MAA).

Concepts:double countingcube geometry

Difficulty rating: 1270

6.

An integer is assigned to each vertex of a cube. The value of an edge is defined to be the sum of the values of the two vertices it touches, and the value of a face is defined to be the sum of the values of the four edges surrounding it. The value of the cube is defined as the sum of the values of its six faces. Suppose the sum of the integers assigned to the vertices is 21.21. What is the value of the cube?

4242

6363

8484

126126

252252

Solution:

Count by incidences. Each edge lies on 22 faces, so the six face values together are 22 times the total of all edge values. Each vertex lies on 33 edges, so the total edge value is 33 times the vertex sum. Chaining these, the cube's value is 2321=126.2 \cdot 3 \cdot 21 = 126. Therefore, the answer is D.

Problem 6 in Other Years