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

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Now it's quite simple to write the inequality as <math>1</math> <math>-</math> <math>\frac{\frac{N}{5}-1}{N+1}</math> <math><</math> <math>\frac{321}{400}</math>
 
Now it's quite simple to write the inequality as <math>1</math> <math>-</math> <math>\frac{\frac{N}{5}-1}{N+1}</math> <math><</math> <math>\frac{321}{400}</math>
 +
 +
We can subtract <math>1</math> on both sides to obtain <math>-</math><math>\frac{\frac{N}{5}-1}{N+1}</math> <math><</math> <math>\frac{-79}{400}</math>
  
 
==See Also==
 
==See Also==

Revision as of 11:48, 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 Solution

Let $N$ $=$ $5$, $P(N)$ $=$ 1


Let $N$ $=$ $10$, $P(N)$ $=$ $\frac{10}{11}$


Let $N$ $=$ $15$, $P(N)$ $=$ $\frac{14}{16}$


Notice that the fraction can be written as $1$ $-$ $\frac{\frac{N}{5}-1}{N+1}$

Now it's quite simple to write the inequality as $1$ $-$ $\frac{\frac{N}{5}-1}{N+1}$ $<$ $\frac{321}{400}$

We can subtract $1$ on both sides to obtain $-$$\frac{\frac{N}{5}-1}{N+1}$ $<$ $\frac{-79}{400}$

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

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