Difference between revisions of "2019 AMC 10A Problems/Problem 4"
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==Solution== | ==Solution== | ||
We want to maximize the number terms. Since we are not limited by positive integers, we know the answer can be greater than 9. We see that the sum of integers from <math>-n</math> to <math>n</math> is <math>0</math>, so we can choose <math>n</math> to be 44. Then the only number to add left is <math>45</math>. | We want to maximize the number terms. Since we are not limited by positive integers, we know the answer can be greater than 9. We see that the sum of integers from <math>-n</math> to <math>n</math> is <math>0</math>, so we can choose <math>n</math> to be 44. Then the only number to add left is <math>45</math>. | ||
| − | Therefore | + | Therefore the number of terms is <math>44+44+1+1 = \boxed{90}</math> |
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| − | < | ||
==See Also== | ==See Also== | ||
Revision as of 18:01, 9 February 2019
- The following problem is from both the 2019 AMC 10A #4 and 2019 AMC 12A #3, so both problems redirect to this page.
Problem
A box contains 
red balls, 
green balls, 
yellow balls, 
blue balls, 
white balls, and 
black balls. What is the minimum number of balls that must be drawn from the box without replacement to guarantee that at least 
balls of a single color will be drawn
Solution
We want to maximize the number terms. Since we are not limited by positive integers, we know the answer can be greater than 9. We see that the sum of integers from 
to 
is 
, so we can choose to be 44. Then the only number to add left is 
.
Therefore the number of terms is 
See Also
| 2019 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 | ||
| 2019 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 2 |
Followed by Problem 4 |
| 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. 
