2020 CIME I Problems/Problem 13

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 $1000$ .

$\frac 23 \cdot h + \frac 25 \cdot a=26$

$h+a=49$

$\frac 23 \cdot (49-a) + \frac 25 \cdot a=26$

$\frac {98}{3}-\frac {2a}{3} + \frac {2a}{5}=26$

$\frac {-4a}{15}= \frac {-20}{3}$

First we thought we should make two variables for two different types of games, $a$ (away games) $h$ (home games). We knew $\frac 23 \cdot h + \frac 25 \cdot a=26$ We also knew that $h+a=49$ , which means $h=49-a$ . So we replaced $h$ in our first equation with $49-a$ , so now it is: $\frac 23 \cdot (49-a) + \frac 25 \cdot a=26$ . Solving this we get: $\frac {98}{3}-\frac {2a}{3} + \frac {2a}{5}=26$ solving this further, we get $\frac {-4a}{15}= \frac {-20}{3}$ . Solving this we get $a=25,$ and going back to $h=49-a,$ we replace $a$ with $25$ and because $49-25=24, h=24$ .

Solution

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2020 CIME I (Problems • Answer Key • Resources)
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2020 CIME I Problems/Problem 13

Problem 13

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