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Difference between revisions of "2016 AMC 10B Problems/Problem 17"

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Now, we can factor even more:
 
Now, we can factor even more:
 
                                 <math>(b+f)(ac+ae+cd+de)</math>
 
                                 <math>(b+f)(ac+ae+cd+de)</math>
−
                                 <math>(b+f)(a(c+e)+d(c+e)</math>
+
                                 <math>(b+f)(a(c+e)+d(c+e))</math>
 
                                 <math>(b+f)(a+d)(c+e)</math>
 
                                 <math>(b+f)(a+d)(c+e)</math>
 
We have the product. Notice how the factors are sums of opposite faces. The best sum for this is to make <math>(7+2)</math>,<math>(6+3)</math>, and <math>(5+4)</math> all factors.
 
We have the product. Notice how the factors are sums of opposite faces. The best sum for this is to make <math>(7+2)</math>,<math>(6+3)</math>, and <math>(5+4)</math> all factors.

Revision as of 17:55, 21 February 2016

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

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: 

                               $abc+abe+bcd+bde+acf+cdf+aef+def$

We notice $b$ is a factor of the first four terms, and $f$ is factor is the last four terms. 

                               $b(ac+ae+cd+de)+f(ac+ae+cd+de)$

Now, we can factor even more: 

                               $(b+f)(ac+ae+cd+de)$
                               $(b+f)(a(c+e)+d(c+e))$
                               $(b+f)(a+d)(c+e)$

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. 

                              $(7+2)(6+3)(5+4)$
                                $9$ $*$   $9$  $*$   $9$
                                   $729$

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

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

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