Difference between revisions of "2013 Mock AIME I Problems/Problem 9"
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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 == | ||
| + | * [[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]] | ||
Revision as of 13:22, 30 July 2024
Problem 9
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, ![\[E_k=\frac{1}{k}+\frac{1}{k}(E_{k-1}+1)+\frac{1}{k}(E_{k-2}+1)+\dots+\frac{1}{k}(E_1+1)=1+\frac{1}{k}(\sum_{i=1}^{k-1} E_i)\]](http://latex.artofproblemsolving.com/b/c/8/bc8e35e3de47d4ea6ade02c84b96345668239991.png)
Doing that calculations gives 
, so 
.
