Difference between revisions of "2013 AMC 10A Problems/Problem 4"
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<math> \textbf{(A)}\ 35 \qquad\textbf{(B)}\ 40 \qquad\textbf{(C)}\ 45 \qquad\textbf{(D)}\ 50 \qquad\textbf{(E)}\ 55 </math> | <math> \textbf{(A)}\ 35 \qquad\textbf{(B)}\ 40 \qquad\textbf{(C)}\ 45 \qquad\textbf{(D)}\ 50 \qquad\textbf{(E)}\ 55 </math> | ||
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| + | 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 1(mod 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: | ||
| + | |||
| + | <math>\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}</math> | ||
| + | |||
| + | The sum of their opponent's scores is 2 + 1 + 4 + 2 + 6 + 3 + 8 + 4 + 10 + 5 = 45, (C) | ||
Revision as of 15:21, 7 February 2013
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?
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 1(mod 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:
The sum of their opponent's scores is 2 + 1 + 4 + 2 + 6 + 3 + 8 + 4 + 10 + 5 = 45, (C)
