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 $1000$.

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


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

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

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

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 $49-25=24, h=24$.


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See also

2020 CIME I (ProblemsAnswer KeyResources)
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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