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

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==Solution 1==
 
==Solution 1==
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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{4n + 2}{5n + 1}</math>, so
+
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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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:
 
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>.
+
<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 its digits is <math>\boxed{\textbf{(A) } 12}</math>.
  
−
==Video Solution==
+
 
 +
==Solution 4==
 +
Draw <math>N</math> balls as described in the problem. First consider the left side arrangements. Notice that for at least <math>\frac{3}{5}N</math> balls must be to the right of the valid left side arrangements. Since we have <math>N+1</math> balls, we know that there are <math>\frac{2}{5}N+1</math> valid left side arrangements. This also goes for the right side arrangements, for 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 its digits is <math>\boxed{\textbf{(A) } 12}</math>.
 +
 
 +
== Video Solution by Pi Academy ==
 +
https://youtu.be/iTkJExTiGkM?si=2dEicIJoLsaeWE-y
 +
 
 +
~ Pi Academy
 +
 
 +
== Video Solution 2==
 
https://youtu.be/YsIPcwuycF0
 
https://youtu.be/YsIPcwuycF0
  
 
~IceMatrix
 
~IceMatrix
−
==Video Solution 2==
+
==Video Solution 3==
  
 
https://www.youtube.com/watch?v=iFSTuWCrosY
 
https://www.youtube.com/watch?v=iFSTuWCrosY

Latest revision as of 18:58, 8 October 2024

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 its digits is $\boxed{\textbf{(A) } 12}$.


Solution 4

Draw $N$ balls as described in the problem. First consider the left side arrangements. Notice that for at least $\frac{3}{5}N$ balls must be to the right of the valid left side arrangements. Since we have $N+1$ balls, we know that there are $\frac{2}{5}N+1$ valid left side arrangements. This also goes for the right side arrangements, for 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 its digits is $\boxed{\textbf{(A) } 12}$.

Video Solution by Pi Academy

https://youtu.be/iTkJExTiGkM?si=2dEicIJoLsaeWE-y

~ Pi Academy

Video Solution 2

https://youtu.be/YsIPcwuycF0

~IceMatrix

Video Solution 3

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

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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