2016 AMC 10B Problems/Problem 17

Revision as of 21:47, 28 January 2017 by Dibjuv (talk | contribs) (Solution)


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 1

Let us call the six sides of our cube $a,b,c,d,e,$ and $f$ (where $a$ is opposite $d$, $c$ is opposite $e$, and $b$ is opposite $f$. Thus, for the eight vertices, we have the following products: $abc$,$abe$,$bcd$,$bde$,$acf$,$cdf$,$cef$, and $def$. Let us find the sum of these products:


We notice $b$ is a factor of the first four terms, and $f$ 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 $(7+2)$,$(6+3)$, and $(5+4)$ all factors.

                                $9$ $*$   $9$  $*$   $9$

Thus our answer is $\textbf{(D)}\ 729$.

Solution 2

We first find the factorization $(b+f)(a+d)(c+e)$ using the method in Solution 1. By using AM-GM, we get, $(b+f)(a+d)(c+e) \le \left( \frac{a+b+c+d+e+f}{3} \right)^3$. To maximize, the factorization, we get the answer is $\left( \frac{27}{3} \right)^3 = \boxed{\textbf{(D)}\ 729}$

Solution 3(Cheap Solution)


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


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). $1680$ is way too high using reasonability after you do this so you put $\boxed{\textbf{D}}$.

See Also

2016 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
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All AMC 10 Problems and Solutions

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