2019 AMC 10A Problem 4

Below is the professionally curated solution for Problem 4 of the 2019 AMC 10A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2019 AMC 10A solutions, or check the answer key.

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Concepts:pigeonhole principleextremal argument

Difficulty rating: 1070

4.

A box contains 2828 red balls, 2020 green balls, 1919 yellow balls, 1313 blue balls, 1111 white balls, and 99 black balls. What is the minimum number of balls that must be drawn from the box without replacement to guarantee that at least 1515 balls of a single color will be drawn?

7575

7676

7979

8484

9191

Solution:

Note that we can pull as many as 1414 balls of each color without ensuring that 1515 balls of one color are drawn.

This means that we can draw all of the black, white and blue balls, along with 1414 red, green, and yellow balls.

This gives us a total of 9+11+13+314 9 + 11 + 13 + 3 \cdot 14 =33+42= 33 + 42 =75.= 75. We need to add one at the end, however, to ensure that we get that 1515th ball of some color, 75+1=76.75 + 1 = 76.

Thus, B is the correct answer.

Problem 4 in Other Years