Difference between revisions of "2019 AMC 10A Problems/Problem 4"
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| + | ==Solution 3== | ||
| + | Pigeon Hole Principle. You want to assume that you have very bad luck. The worst the choosing socks could go would be choosing 14 red balls, 14 green, 14 yellow, 13 blue, 11 white, and 9 black. These add up to 75, now no matter what color you choose it makes 15. (There are no more blue, white, and black socks left.) | ||
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==See Also== | ==See Also== | ||
Revision as of 20:46, 6 January 2021
- 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
By choosing the maximum number of balls while getting 
of each color, we could have chosen 
red balls, green balls, yellow balls, blue balls, white balls, and black balls, for a total of 
balls. Picking one more ball guarantees that we will get balls of a color -- either red, green, or yellow. Thus the answer is 
.
Video Solution 1
Education, The Study of Everything
Video Solution 2
~savannahsolver
Solution 3
Pigeon Hole Principle. You want to assume that you have very bad luck. The worst the choosing socks could go would be choosing 14 red balls, 14 green, 14 yellow, 13 blue, 11 white, and 9 black. These add up to 75, now no matter what color you choose it makes 15. (There are no more blue, white, and black socks left.)
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. 
