# 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

The problems on this page are copyrighted by the MAC's Christmas Mathematics Competitions. 