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> | ||
| − | ==Solution== | + | ==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 <math>2</math>. Thus, from this, we know that the five games where they lost by one run were when they scored <math>1</math>, <math>3</math>, <math>5</math>, <math>7</math>, and <math>9</math> runs, and the others are where they scored twice as many runs. We can make the following chart: | 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 <math>2</math>. Thus, from this, we know that the five games where they lost by one run were when they scored <math>1</math>, <math>3</math>, <math>5</math>, <math>7</math>, and <math>9</math> runs, and the others are where they scored twice as many runs. We can make the following chart: | ||
| Line 26: | Line 26: | ||
The sum of their opponent's scores is <math>2 + 1 + 4 + 2 + 6 + 3 + 8 + 4 + 10 + 5 = \boxed{\textbf{(C) }45}</math> | The sum of their opponent's scores is <math>2 + 1 + 4 + 2 + 6 + 3 + 8 + 4 + 10 + 5 = \boxed{\textbf{(C) }45}</math> | ||
| + | |||
| + | ==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, <math>1,2,3,4,5</math> when the team has even runs. The opponents will have an arithmetic sequence of even runs, <math>2,4,6,8,10</math>, when the team has odd runs. The sum of their runs is <math>3\cdot(1+2+3+4+5)=3\cdot15=45=\fbox{C}</math> | ||
| + | ~dragnin | ||
| + | |||
| + | ==Video Solution (CREATIVE THINKING)== | ||
| + | https://youtu.be/Z6SSR22FIWg | ||
| + | |||
| + | ~Education, the Study of Everything | ||
| + | |||
==Video Solution== | ==Video Solution== | ||
Latest revision as of 20:40, 4 July 2023
Contents
Problem
A softball team played ten games, scoring 
, 
, 
, 
, 
, 
, 
, 
, 
, and 
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?
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 . Thus, from this, we know that the five games where they lost by one run were when they scored , , , , and 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 
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, 
when the team has even runs. The opponents will have an arithmetic sequence of even runs, 
, when the team has odd runs. The sum of their runs is 
~dragnin
Video Solution (CREATIVE THINKING)
~Education, the Study of Everything
Video Solution
https://www.youtube.com/watch?v=2vf843cvVzo?t=91
~sugar_rush
See Also
| 2013 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 3 |
Followed by Problem 5 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
| 2013 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 1 |
Followed by Problem 3 |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
| All AMC 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
