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Difference between revisions of "2016 AMC 10A Problems/Problem 18"

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Let <math>n = \frac{N}{5}</math>. Then, consider <math>5</math> blocks of <math>n</math> green balls in a line, along with the red ball. Shuffling the line is equivalent to choosing one of the <math>N + 1</math> positions between the green balls to insert the red ball. Less than <math>\frac{3}{5}</math> of the green balls will be on the same side of the red ball if the red ball is inserted in the middle block of <math>n</math> balls, and there are <math>n - 1</math> positions where this happens. Thus, <math>P(N) = 1 - \frac{n - 1}{N + 1} = \frac{0.8N + 2}{N + 1}</math>.
 
Let <math>n = \frac{N}{5}</math>. Then, consider <math>5</math> blocks of <math>n</math> green balls in a line, along with the red ball. Shuffling the line is equivalent to choosing one of the <math>N + 1</math> positions between the green balls to insert the red ball. Less than <math>\frac{3}{5}</math> of the green balls will be on the same side of the red ball if the red ball is inserted in the middle block of <math>n</math> balls, and there are <math>n - 1</math> positions where this happens. Thus, <math>P(N) = 1 - \frac{n - 1}{N + 1} = \frac{0.8N + 2}{N + 1}</math>.
 
Solving the inequality <math>P(N) = \frac{0.8N + 2}{N + 1} < \frac{321}{400}</math> gives <math>N > 479</math>, so the least value of <math>N</math> is <math>480</math>. The sum of the digits of <math>480</math> is <math>\boxed{12}</math>.
 
Solving the inequality <math>P(N) = \frac{0.8N + 2}{N + 1} < \frac{321}{400}</math> gives <math>N > 479</math>, so the least value of <math>N</math> is <math>480</math>. The sum of the digits of <math>480</math> is <math>\boxed{12}</math>.
 +
 +
<asy>
 +
draw(circle((0,0),3));
 +
draw(circle((0,0),1));
 +
draw(circle((1,sqrt(3)),1));
 +
draw(circle((-1,sqrt(3)),1));
 +
draw(circle((-1,-sqrt(3)),1)); draw(circle((1,-sqrt(3)),1));
 +
draw(circle((2,0),1)); draw(circle((-2,0),1)); </asy>
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2016|ab=A|num-b=17|num-a=19}}
 
{{AMC10 box|year=2016|ab=A|num-b=17|num-a=19}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 23:50, 3 February 2016

Each vertex of a cube is to be labeled with an integer $1$ through $8$, with each integer being used once, in such a way that the sum of the four numbers on the vertices of a face is the same for each face. Arrangements that can be obtained from each other through rotations of the cube are considered to be the same. How many different arrangements are possible?

$\textbf{(A) } 1\qquad\textbf{(B) } 3\qquad\textbf{(C) }6 \qquad\textbf{(D) }12 \qquad\textbf{(E) }24$

Solution

Let $n = \frac{N}{5}$. Then, consider $5$ blocks of $n$ green balls in a line, along with the red ball. Shuffling the line is equivalent to choosing one of the $N + 1$ positions between the green balls to insert the red ball. Less than $\frac{3}{5}$ of the green balls will be on the same side of the red ball if the red ball is inserted in the middle block of $n$ balls, and there are $n - 1$ positions where this happens. Thus, $P(N) = 1 - \frac{n - 1}{N + 1} = \frac{0.8N + 2}{N + 1}$. Solving the inequality $P(N) = \frac{0.8N + 2}{N + 1} < \frac{321}{400}$ gives $N > 479$, so the least value of $N$ is $480$. The sum of the digits of $480$ is $\boxed{12}$.

[asy] draw(circle((0,0),3)); draw(circle((0,0),1)); draw(circle((1,sqrt(3)),1)); draw(circle((-1,sqrt(3)),1)); draw(circle((-1,-sqrt(3)),1)); draw(circle((1,-sqrt(3)),1)); draw(circle((2,0),1)); draw(circle((-2,0),1)); [/asy]

See Also

2016 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 17 		Followed by
Problem 19
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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