Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
(→‎Solution)
Line 25: Line 25:
  
 
==Pattern==
 
==Pattern==
−
Let N = <math>5</math>, p(N) = 1
+
Let N = <math>5</math>, P(N) = 1
−
Let N = <math>10</math>, p(N) = \frac{1}{11}<math>
+
Let N = <math>10</math>, P(N) = \frac{1}{11}<math>
−
Let N = </math>15<math>, notice that p(n) = \frac{2}{16}</math>
+
Let N = </math>15<math>, P(N) = \frac{2}{16}</math>
−
So as we notice, starting from n = <math>10</math>, p(n) can be written in the form \frac{x}{N+1}$, where x starts at 1 when N = 10, and increases 1 every multiple of 5.
+
So as we notice, starting from N = <math>10</math>, P(N) can be written in the form \frac{x}{N+1}$, where x starts at 1 when N = 10, and increases 1 every multiple of 5.
  
 
==See Also==
 
==See Also==

Revision as of 11:01, 20 May 2016

Problem

Let $N$ be a positive multiple of $5$. One red ball and $N$ green balls are arranged in a line in random order. Let $P(N)$ be the probability that at least $\tfrac{3}{5}$ of the green balls are on the same side of the red ball. Observe that $P(5)=1$ and that $P(N)$ approaches $\tfrac{4}{5}$ as $N$ grows large. What is the sum of the digits of the least value of $N$ such that $P(N) < \tfrac{321}{400}$?

$\textbf{(A) } 12 \qquad \textbf{(B) } 14 \qquad \textbf{(C) }16 \qquad \textbf{(D) } 18 \qquad \textbf{(E) } 20$

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, $P(N) = 1 - \frac{n - 1}{N + 1} = \frac{4n + 2}{5n + 1}$, so

\[P(N) = \frac{4n + 2}{5n + 1} < \frac{321}{400}.\]

Multiplying both sides of the inequality by $400(5n+1)$, we have

\[400(4n+2)<321(5n+1),\]

and by the distributive property,

\[1600n+800<1605n+321.\]

Subtracting $1600n+321$ on both sides of the inequality gives us

\[479<5n.\]

Therefore, $N=5n>479$, so the least possible value of $N$ is $480$. The sum of the digits of $480$ is $\boxed{\textbf{(A) } 12}$.

Pattern

Let N = $5$, P(N) = 1 Let N = $10$, P(N) = \frac{1}{11}$Let N =$15$, P(N) = \frac{2}{16}$ So as we notice, starting from N = $10$, P(N) can be written in the form \frac{x}{N+1}$, where x starts at 1 when N = 10, and increases 1 every multiple of 5.

See Also

2016 AMC 10A (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
2016 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10A_Problems/Problem_17&oldid=78622"