Difference between revisions of "2019 AMC 10A Problems/Problem 4"
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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 we have: | Therefore we have: | ||
− | < | + | |
+ | <cmath>\begin{center} 44+44+1+1 = \boxed{90} \end{center}</cmath> | ||
==See Also== | ==See Also== |
Revision as of 17:59, 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 we have:
\[\begin{center} 44+44+1+1 = \boxed{90} \end{center}\] (Error compiling LaTeX. Unknown error_msg)
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.