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== | ||
Revision as of 22:46, 6 July 2022
Problem 13
Chris writes on a piece of paper the positive integers from 
to 
in that order. Then, he randomly writes either 
or 
between every two adjacent numbers, each with equal probability. The expected value of the expression he writes can be expressed as 
for relatively prime positive integers 
and 
. Find the remainder when 
is divided by 
.
First we thought we should make two variables for two different types of games, 
(away games) 
(home games). We knew We also knew that , which means 
. So we replaced in our first equation with 
, so now it is: . Solving this we get: solving this further, we get . Solving this we get 
and going back to 
we replace with 
and because 
.
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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