Difference between revisions of "2013 Mock AIME I Problems/Problem 9"
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− | ==Problem | + | ==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>. | ||
==Solution== | ==Solution== | ||
− | 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} 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>. |
+ | |||
+ | == See also == | ||
+ | * [[2013 Mock AIME I Problems]] | ||
+ | * [[2013 Mock AIME I Problems/Problem 8|Preceded by Problem 8]] | ||
+ | * [[2013 Mock AIME I Problems/Problem 10|Followed by Problem 10]] |
Latest revision as of 14:31, 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 is the expected number of turns it takes to short circuit all of the lights, find
.
Solution
Let be the expected value of turns it takes to short circuit
lights. Note that
,
, and in general,
Doing that calculations gives
, so
.