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

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<math>\textbf{(A) } 12 \qquad \textbf{(B) } 14 \qquad \textbf{(C) }16 \qquad \textbf{(D) } 18 \qquad \textbf{(E) } 20</math>
 
<math>\textbf{(A) } 12 \qquad \textbf{(B) } 14 \qquad \textbf{(C) }16 \qquad \textbf{(D) } 18 \qquad \textbf{(E) } 20</math>
  
==Solution==
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==Solution 1==
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{4n + 2}{5n + 1}</math>, so
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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 inside 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{4n + 2}{5n + 1}</math>, so
  
 
<cmath>P(N) = \frac{4n + 2}{5n + 1} < \frac{321}{400}.</cmath>
 
<cmath>P(N) = \frac{4n + 2}{5n + 1} < \frac{321}{400}.</cmath>
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Therefore, <math>N=5n>479</math>, so the least possible value of <math>N</math> is <math>480</math>. The sum of the digits of <math>480</math> is <math>\boxed{\textbf{(A) } 12}</math>.
 
Therefore, <math>N=5n>479</math>, so the least possible value of <math>N</math> is <math>480</math>. The sum of the digits of <math>480</math> is <math>\boxed{\textbf{(A) } 12}</math>.
  
==Pattern Solution==
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==Solution 2 (Pattern Solution)==
Let <math>N</math> <math>=</math> <math>5</math>, <math>P(N)</math> <math>=</math> 1
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Let <math>N=5</math>, <math>P(N)=1</math> (Given)
  
  
Let <math>N</math> <math>=</math> <math>10</math>, <math>P(N)</math> <math>=</math> <math>\frac{10}{11}</math>
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Let <math>N=10</math>, <math>P(N)=\frac{10}{11}</math>
  
  
Let <math>N</math> <math>=</math> <math>15</math>, <math>P(N)</math> <math>=</math> <math>\frac{14}{16}</math>
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Let <math>N=15</math>, <math>P(N)=\frac{14}{16}</math>
  
  
Notice that the fraction can be written as <math>1</math> <math>-</math> <math>\frac{\frac{N}{5}-1}{N+1}</math>
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Notice that the fraction can be written as <math>1-\frac{\frac{N}{5}-1}{N+1}</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>
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Now it's quite simple to write the inequality as <math>1-\frac{\frac{N}{5}-1}{N+1}<\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>-</math><math>\frac{79}{400}</math>
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We can subtract <math>1</math> on both sides to obtain <math>-\frac{\frac{N}{5}-1}{N+1}<-\frac{79}{400}</math>
  
Dividing both sides by <math>-1</math>, we derive <math>\frac{\frac{N}{5}-1}{N+1}</math> <math>></math> <math>\frac{79}{400}</math>. (Switch the inequality sign when dividing by <math>-1</math>)
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Dividing both sides by <math>-1</math>, we derive <math>\frac{\frac{N}{5}-1}{N+1}>\frac{79}{400}</math>. (Switch the inequality sign when dividing by <math>-1</math>)
  
 
We then cross multiply to get <math>80N - 400 > 79N + 79</math>
 
We then cross multiply to get <math>80N - 400 > 79N + 79</math>
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Finally we get <math>N > 479</math>
 
Finally we get <math>N > 479</math>
  
To get <math>N = 480</math>
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To achieve <math>N = 480</math>
  
So the sum of the digits of <math>N</math> = <math>\boxed{\textbf{(A) } 12}</math>.
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So the sum of the digits of <math>N</math> = <math>\boxed{\textbf{(A) } 12}</math>
  
<math>solution</math> <math>by</math> <math>AOPS12142015 (Kyle L.)</math>
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==Solution 3==
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We are trying to find the number of places to put the red ball, such that <math>\frac{3}{5}</math> of the green balls or more are on one side of it. Notice that we can put the ball in a number of spaces describable with <math>N</math>: Trying a few values, we see that the ball "works" in places <math>1</math> to <math>\frac{2}{5}N + 1</math> and spaces <math>\frac{3}{5}N+1 </math> to <math>N+1</math>. This is a total of <math>\frac{4}{5}N + 2</math> spaces, over a total possible <math>N + 1</math> places to put the ball. So:
 +
 
 +
<math>\frac{\frac{4}{5}N + 2}{N+1} = \frac{321}{400} \rightarrow N = 479.</math> And we know that the next value is what we are looking for, so <math>N+1 = 480</math>, and the sum of it's digits is <math>\boxed{\textbf{(A) } 12}</math>.
 +
 
 +
==Video Solution==
 +
https://youtu.be/YsIPcwuycF0
 +
 
 +
~IceMatrix
 +
==Video Solution 2==
 +
 
 +
https://www.youtube.com/watch?v=iFSTuWCrosY
  
 
==See Also==
 
==See Also==

Revision as of 22:48, 26 September 2021

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 1

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 inside 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 $\boxed{\textbf{(A) } 12}$.

Solution 2 (Pattern Solution)

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


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 $-\frac{\frac{N}{5}-1}{N+1}<-\frac{79}{400}$

Dividing both sides by $-1$, we derive $\frac{\frac{N}{5}-1}{N+1}>\frac{79}{400}$. (Switch the inequality sign when dividing by $-1$)

We then cross multiply to get $80N - 400 > 79N + 79$

Finally we get $N > 479$

To achieve $N = 480$

So the sum of the digits of $N$ = $\boxed{\textbf{(A) } 12}$

Solution 3

We are trying to find the number of places to put the red ball, such that $\frac{3}{5}$ of the green balls or more are on one side of it. Notice that we can put the ball in a number of spaces describable with $N$: Trying a few values, we see that the ball "works" in places $1$ to $\frac{2}{5}N + 1$ and spaces $\frac{3}{5}N+1$ to $N+1$. This is a total of $\frac{4}{5}N + 2$ spaces, over a total possible $N + 1$ places to put the ball. So:

$\frac{\frac{4}{5}N + 2}{N+1} = \frac{321}{400} \rightarrow N = 479.$ And we know that the next value is what we are looking for, so $N+1 = 480$, and the sum of it's digits is $\boxed{\textbf{(A) } 12}$.

Video Solution

https://youtu.be/YsIPcwuycF0

~IceMatrix

Video Solution 2

https://www.youtube.com/watch?v=iFSTuWCrosY

See Also

2016 AMC 10A (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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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