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Difference between revisions of "2013 Mock AIME I Problems/Problem 9"

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==Problem 9==
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==Problem==
 
In a magic circuit, there are six lights in a series, and if one of the lights short circuit, then all lights after it will short circuit as well, without affecting the lights before it. Once a turn, a random light that isn’t already short circuited is short circuited. If <math> E </math> is the expected number of turns it takes to short circuit all of the lights, find <math> 100E </math>.
 
In a magic circuit, there are six lights in a series, and if one of the lights short circuit, then all lights after it will short circuit as well, without affecting the lights before it. Once a turn, a random light that isn’t already short circuited is short circuited. If <math> E </math> is the expected number of turns it takes to short circuit all of the lights, find <math> 100E </math>.
  
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Let <math>E_n</math> be the expected value of turns it takes to short circuit <math>n</math> lights. Note that <math>E_1=1</math>, <math>E_2=\frac{1}{2}+\frac{1}{2}(E_1+1)</math>, and in general, <cmath>E_k=\frac{1}{k}+\frac{1}{k}(E_{k-1}+1)+\frac{1}{k}(E_{k-2}+1)+\dots+\frac{1}{k}(E_1+1)=1+\frac{1}{k}(\sum_{i=1}^{k-1} E_i)</cmath> Doing that calculations gives <math>E_6=\frac{49}{20}</math>, so <math>100E_6=\boxed{245}</math>.
 
Let <math>E_n</math> be the expected value of turns it takes to short circuit <math>n</math> lights. Note that <math>E_1=1</math>, <math>E_2=\frac{1}{2}+\frac{1}{2}(E_1+1)</math>, and in general, <cmath>E_k=\frac{1}{k}+\frac{1}{k}(E_{k-1}+1)+\frac{1}{k}(E_{k-2}+1)+\dots+\frac{1}{k}(E_1+1)=1+\frac{1}{k}(\sum_{i=1}^{k-1} E_i)</cmath> Doing that calculations gives <math>E_6=\frac{49}{20}</math>, so <math>100E_6=\boxed{245}</math>.
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== See also ==
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* [[2013 Mock AIME I Problems]]
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* [[2013 Mock AIME I Problems/Problem 8|Preceded by Problem 8]]
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* [[2013 Mock AIME I Problems/Problem 10|Followed by Problem 10]]

Latest revision as of 13:31, 30 July 2024

Problem

In a magic circuit, there are six lights in a series, and if one of the lights short circuit, then all lights after it will short circuit as well, without affecting the lights before it. Once a turn, a random light that isn’t already short circuited is short circuited. If $E$ is the expected number of turns it takes to short circuit all of the lights, find $100E$.

Solution

Let $E_n$ be the expected value of turns it takes to short circuit $n$ lights. Note that $E_1=1$, $E_2=\frac{1}{2}+\frac{1}{2}(E_1+1)$, and in general, \[E_k=\frac{1}{k}+\frac{1}{k}(E_{k-1}+1)+\frac{1}{k}(E_{k-2}+1)+\dots+\frac{1}{k}(E_1+1)=1+\frac{1}{k}(\sum_{i=1}^{k-1} E_i)\] Doing that calculations gives $E_6=\frac{49}{20}$, so $100E_6=\boxed{245}$.

See also

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