2016 AMC 12B Problems/Problem 15

Revision as of 11:09, 21 February 2016 by Mathmaster2012 (talk | contribs) (Created page with "==Problem== All the numbers <math>2, 3, 4, 5, 6, 7</math> are assigned to the six faces of a cube, one number to each face. For each of the eight vertices of the cube, a prod...")
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

Problem

All the numbers $2, 3, 4, 5, 6, 7$ are assigned to the six faces of a cube, one number to each face. For each of the eight vertices of the cube, a product of three numbers is computed, where the three numbers are the numbers assigned to the three faces that include that vertex. What is the greatest possible value of the sum of these eight products?

$\textbf{(A)}\ 312 \qquad \textbf{(B)}\ 343 \qquad \textbf{(C)}\ 625 \qquad \textbf{(D)}\ 729 \qquad \textbf{(E)}\ 1680$

Solution

See Also

2016 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 16
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png