Difference between revisions of "2019 AMC 10A Problems/Problem 4"
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Namely, we can draw up to <math>14</math> red balls, <math>14</math> green balls, <math>14</math> yellow balls, <math>13</math> blue balls, <math>11</math> white balls, and <math>9</math> black balls, for a total of <math>75</math> balls, without drawing <math>15</math> balls of any one color. Drawing one more ball guarantees that we will get <math>15</math> balls of one color — either red, green, or yellow. Thus, the answer is <math>75 + 1 = \boxed{\textbf{(B) } 76}</math>. | Namely, we can draw up to <math>14</math> red balls, <math>14</math> green balls, <math>14</math> yellow balls, <math>13</math> blue balls, <math>11</math> white balls, and <math>9</math> black balls, for a total of <math>75</math> balls, without drawing <math>15</math> balls of any one color. Drawing one more ball guarantees that we will get <math>15</math> balls of one color — either red, green, or yellow. Thus, the answer is <math>75 + 1 = \boxed{\textbf{(B) } 76}</math>. | ||
| − | ==Solution ( | + | ==Solution 2 (cheese)== |
| + | After a through look through the answer choices we see that 75 and 76 are only one from each other, AoPs may have predicted that some people might silly this problem and pick 75 after seeing that it was an answer choice. Also accounting the fact that this is problem 4 you can relatively safely pick 76 immediately after seeing the answer choices. | ||
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| + | Note: but, we would need to know that it's possible to silly and choose the answer choice of 75 after adding 14(3)+13+11+9=75 and forgetting to add 1. Thus, better to make the quick calculation rather than use a cheese. | ||
| + | |||
| + | ~yvz2900 | ||
==Video Solution 1== | ==Video Solution 1== | ||
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{{AMC12 box|year=2019|ab=A|num-b=2|num-a=4}} | {{AMC12 box|year=2019|ab=A|num-b=2|num-a=4}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
| + | [[Category: Logic Problems]] | ||
Latest revision as of 15:07, 20 October 2025
- The following problem is from both the 2019 AMC 10A #4 and 2019 AMC 12A #3, so both problems redirect to this page.
Contents
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 try to find the worst case scenario where we can find the maximum number of balls that can be drawn while getting 
of each color by applying the pigeonhole principle and through this we get a perfect guarantee.
Namely, we can draw up to 
red balls, green balls, yellow balls, blue balls, white balls, and black balls, for a total of 
balls, without drawing balls of any one color. Drawing one more ball guarantees that we will get balls of one color — either red, green, or yellow. Thus, the answer is 
.
Solution 2 (cheese)
After a through look through the answer choices we see that 75 and 76 are only one from each other, AoPs may have predicted that some people might silly this problem and pick 75 after seeing that it was an answer choice. Also accounting the fact that this is problem 4 you can relatively safely pick 76 immediately after seeing the answer choices.
Note: but, we would need to know that it's possible to silly and choose the answer choice of 75 after adding 14(3)+13+11+9=75 and forgetting to add 1. Thus, better to make the quick calculation rather than use a cheese.
~yvz2900
Video Solution 1
Education, The Study of Everything
Video Solution 2
https://youtu.be/8WrdYLw9_ns?t=23
~ pi_is_3.14
Video Solution 3
~savannahsolver
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 | |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
