2016 AMC 10B Problems/Problem 17
Contents
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 1
Let us call the six sides of our cube and (where is opposite , is opposite , and is opposite . Thus, for the eight vertices, we have the following products: ,,,,,,, and . Let us find the sum of these products:
We notice is a factor of the first four terms, and is factor is the last four terms.
Now, we can factor even more:
We have the product. Notice how the factors are sums of opposite faces. The best sum for this is to make ,, and all factors.
Thus our answer is .
Solution 2
We first find the factorization using the method in Solution 1. By using AM-GM, we get, . To maximize, the factorization, we get the answer is
Solution 3 (Cheap Solution)
Warning
Only use this if you are stuck on the problem or are low on time, and if you don't want to get the correct answer.
Solution
Create a pairing that seems to intuitively seem to be optimal value. So put a number and it's complement(the number that's the difference of 9 and this number). is way too high using reasonability after you do this so you put .
See Also
2016 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 16 |
Followed by Problem 18 | |
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 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.