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Difference between revisions of "2020 CIME I Problems/Problem 13"

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==Problem 13==
 
==Problem 13==
 
Chris writes on a piece of paper the positive integers from <math>1</math> to <math>8</math> in that order. Then, he randomly writes either <math>+</math> or <math>\times</math> between every two adjacent numbers, each with equal probability. The expected value of the expression he writes can be expressed as <math>\frac{p}{q}</math> for relatively prime positive integers <math>p</math> and <math>q</math>. Find the remainder when <math>p+q</math> is divided by <math>1000</math>.
 
Chris writes on a piece of paper the positive integers from <math>1</math> to <math>8</math> in that order. Then, he randomly writes either <math>+</math> or <math>\times</math> between every two adjacent numbers, each with equal probability. The expected value of the expression he writes can be expressed as <math>\frac{p}{q}</math> for relatively prime positive integers <math>p</math> and <math>q</math>. Find the remainder when <math>p+q</math> is divided by <math>1000</math>.
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<math>\frac 23 \cdot h + \frac 25 \cdot a=26</math>
 
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<math>h+a=49</math>
 
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<math>\frac 23 \cdot (49-a) + \frac 25 \cdot a=26</math>
 
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<math>\frac {98}{3}-\frac {2a}{3} + \frac {2a}{5}=26</math>
 
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<math>\frac {-4a}{15}= \frac {-20}{3}</math>
 
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First we thought we should make two variables for two different types of games, <math>a</math> (away games) <math>h</math> (home games). We knew <math>\frac 23 \cdot h + \frac 25 \cdot a=26</math> We also knew that <math>h+a=49</math>, which means <math>h=49-a</math>. So we replaced <math>h</math> in our first equation with <math>49-a</math>, so now it is: <math>\frac 23 \cdot (49-a) + \frac 25 \cdot a=26</math>. Solving this we get: <math>\frac {98}{3}-\frac {2a}{3} + \frac {2a}{5}=26</math> solving this further, we get <math>\frac {-4a}{15}= \frac {-20}{3}</math>. Solving this we get <math>a=25,</math> and going back to <math>h=49-a,</math> we replace <math>a</math> with <math>25</math> and because <math>49-25=24, h=24</math>.
 
  
 
==Solution==
 
==Solution==

Latest revision as of 13:51, 22 December 2023

Problem 13

Chris writes on a piece of paper the positive integers from $1$ to $8$ in that order. Then, he randomly writes either $+$ or $\times$ between every two adjacent numbers, each with equal probability. The expected value of the expression he writes can be expressed as $\frac{p}{q}$ for relatively prime positive integers $p$ and $q$. Find the remainder when $p+q$ is divided by $1000$.

Solution

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See also

2020 CIME I (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
All CIME Problems and Solutions

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