Difference between revisions of "2016 AMC 12B Problems/Problem 15"

Revision as of 00:25, 4 August 2023

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 1

First assign each face the letters $a,b,c,d,e,f$ . The sum of the product of the faces is $abc+acd+ade+aeb+fbc+fcd+fde+feb$ . We can factor this into $(a+f)(b+c)(d+e)$ which is the product of the sum of each pair of opposite faces. In order to maximize $(a+f)(b+c)(d+e)$ we use the numbers $(7+2)(6+3)(5+4)$ or $\boxed{\textbf{(D)}\ 729 }$ .

Solution 2

We'll proceed from the factoring process above. By the AM-GM inequality,

$\frac{a_1+a_2+a_3}{3}\geq\sqrt[3]{a_1a_2a_3}$

Cubing both sides,

$\left(\frac{a_1+a_2+a_3}{3}\right)^3\geq{a_1a_2a_3}$

Let $a_1=(a+f)$ , $a_2=(b+c)$ , and $a_3=(d+e)$ . Let's substitute in these values.

$\left(\frac{a+b+c+d+e+f}{3}\right)^3\geq{(a+f)(b+c)(d+e)}$

$a+b+c+d+e+f$ is fixed at 27.

$\left(\frac{27}{3}\right)^3\geq{(a+f)(b+c)(d+e)}$

$\boxed{\textbf{(D)}\ 729 }\geq{(a+f)(b+c)(d+e)}$

Solution 3 (really fast)

First, we see that we want to try and maximize each vertex. Since the multiplication of each vertex is the product of three values, we want to maximize those three values. Doing so, we see that we want them to be as close as possible, giving $4.5^3$ (the average of all the values). This gives us the maximum for each vertex, multiplied by the 8 vertices, yields our answer $\boxed{\textbf{(D)}\ 729 }$ Also note that if you cannot evaluate $4.5^3$ quickly, a rough approximation of $5*4.5*4*8$ will yield 720, very close to our answer. -rayprati

Note about note above: $4.5=\frac{9}{2}.$ So, we want to evaluate $(\frac{9}{2})^3 \cdot 8 = \frac{729}{8} \cdot 8 = 729.$ So, it’s quite easy to evaluate ~solasky

Solution 4

It is obvious to put $5$ , $6$ , and $7$ on the faces that share the same vertex. As $4$ is the next biggest number, the face with $4$ has to be next to the faces with $6$ and $7$ . As $4$ is the next biggest number, the face with $3$ has to be next to the faces with $5$ and $7$ . making the face with $2$ next to the faces with $6$ and $5$ .

Therefore the answer is $7 \cdot 5 \cdot 6 + 7 \cdot 4 \cdot 6 + 7 \cdot 3 \cdot 5 + 7 \cdot 3 \cdot 4 + 2 \cdot 5 \cdot 6 + 2 \cdot 4 \cdot 6 + 2 \cdot 3 \cdot 5 + 2 \cdot 3 \cdot 4 = \boxed{\textbf{(D)}\ 729 }$

Note: 1680 is impossible to achieve as it requires $210 = 7 \cdot 6 \cdot 5$ for each vertex.

~isabelchen

@@ Line 13: / Line 13: @@
 ==Solution 2==
-We proceed from the factorization in the above solution. By the AM-GM inequality,
+We'll proceed from the factoring process above.
+By the AM-GM inequality,
 <cmath>\frac{a_1+a_2+a_3}{3}\geq\sqrt[3]{a_1a_2a_3}</cmath>
@@ Line 32: / Line 33: @@
 ==Solution 3 (really fast)==
-First, we see that we want to try and maximize each vertex. Since the multiplication of each vertex is the product of three values, we want to maximize those three values. Doing so, we see that we want them to be as close as possible, giving <math>4.5^3</math> (the average of all the values). This gives us the maximum for each vertex, multiplied by the 8 vertices, yields our answer <math>\boxed{\textbf{(D)}\ 729 }</math>. Also note that if you cannot evaluate <math>4.5^3</math> quickly, a rough approximation of <math>5*4.5*4*8</math> will yield 720, very close to our answer.
+First, we see that we want to try and maximize each vertex. Since the multiplication of each vertex is the product of three values, we want to maximize those three values. Doing so, we see that we want them to be as close as possible, giving <math>4.5^3</math> (the average of all the values). This gives us the maximum for each vertex, multiplied by the 8 vertices, yields our answer <math>\boxed{\textbf{(D)}\ 729 }</math> Also note that if you cannot evaluate <math>4.5^3</math> quickly, a rough approximation of <math>5*4.5*4*8</math> will yield 720, very close to our answer. -rayprati
+Note about note above: <math>4.5=\frac{9}{2}.</math> So, we want to evaluate <math>(\frac{9}{2})^3 \cdot 8 = \frac{729}{8} \cdot 8 = 729.</math> So, it’s quite easy to evaluate ~solasky
+==Solution 4==
+It is obvious to put <math>5</math>, <math>6</math>, and <math>7</math> on the faces that share the same vertex. As <math>4</math> is the next biggest number, the face with <math>4</math> has to be next to the faces with <math>6</math> and <math>7</math>. As <math>4</math> is the next biggest number, the face with <math>3</math> has to be next to the faces with <math>5</math> and <math>7</math>. making the face with <math>2</math> next to the faces with <math>6</math> and <math>5</math>.
+Therefore the answer is <math>7 \cdot 5 \cdot 6 + 7 \cdot 4 \cdot 6 + 7 \cdot 3 \cdot 5 + 7 \cdot  3 \cdot 4 + 2 \cdot 5 \cdot 6 + 2 \cdot 4 \cdot 6 + 2 \cdot 3 \cdot 5 + 2 \cdot  3 \cdot 4 = \boxed{\textbf{(D)}\ 729 }</math>
+Note: 1680 is impossible to achieve as it requires <math>210 = 7 \cdot 6 \cdot 5</math> for each vertex.
+~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]
 ==See Also==
 {{AMC12 box|year=2016|ab=B|num-b=14|num-a=16}}
 {{MAA Notice}}

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

Page

Toolbox

Search