# 2013 AMC 10A Problems/Problem 4

## Problem

A softball team played ten games, scoring $1$, $2$, $3$, $4$, $5$, $6$, $7$, $8$, $9$, and $10$ runs. They lost by one run in exactly five games. In each of their other games, they scored twice as many runs as their opponent. How many total runs did their opponents score? $\textbf{(A)}\ 35 \qquad\textbf{(B)}\ 40 \qquad\textbf{(C)}\ 45 \qquad\textbf{(D)}\ 50 \qquad\textbf{(E)}\ 55$

## Solution 1

We know that, for the games where they scored an odd number of runs, they cannot have scored twice as many runs as their opponents, as odd numbers are not divisible by $2$. Thus, from this, we know that the five games where they lost by one run were when they scored $1$, $3$, $5$, $7$, and $9$ runs, and the others are where they scored twice as many runs. We can make the following chart: $\begin{tabular}{|l|l|} \hline Them & Opponent \\ \hline 1 & 2 \\ 2 & 1 \\ 3 & 4 \\ 4 & 2 \\ 5 & 6 \\ 6 & 3 \\ 7 & 8 \\ 8 & 4 \\ 9 & 10 \\ 10 & 5 \\ \hline \end{tabular}$

The sum of their opponent's scores is $2 + 1 + 4 + 2 + 6 + 3 + 8 + 4 + 10 + 5 = \boxed{\textbf{(C) }45}$

## Solution 2 (patterns and easier arithmetic)

The team must've won the games with the even runs and lost the ones with the odd runs. The opponents will have an arithmetic sequence of runs, $1,2,3,4,5$ when the team has even runs. The opponents will have an arithmetic sequence of even runs, $2,4,6,8,10$, when the team has odd runs. The sum of their runs is $3\cdot(1+2+3+4+5)=3\cdot15=45=\fbox{C}$ ~dragnin

## Video Solution (CREATIVE THINKING)

~Education, the Study of Everything

## Video Solution

~sugar_rush

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 