Difference between revisions of "2016 AMC 12B Problems/Problem 15"
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==Solution== | ==Solution== | ||
| + | First assign each face the letters <math>a,b,c,d,e,f</math>. The sum of the product of the faces is <math>abc+acd+ade+aeb+fbc+fcd+fde+feb</math>. We can factor this into <math>(a+f)(b+c)(d+e)</math> which is the product of the sum of each pair of opposite faces. In order to <math>(a+f)(b+c)(d+e)</math> we use the numbers <math>(7+2)(6+3)(5+4)</math> or <math>\boxed{729}</math> | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2016|ab=B|num-b=14|num-a=16}} | {{AMC12 box|year=2016|ab=B|num-b=14|num-a=16}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 16:39, 21 February 2016
Problem
All the numbers 
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?
Solution
First assign each face the letters 
. The sum of the product of the faces is 
. We can factor this into 
which is the product of the sum of each pair of opposite faces. In order to we use the numbers 
or 
See Also
| 2016 AMC 12B (Problems • Answer Key • Resources) | |
| 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. 
