1994 AMC 8 Problem 21

Below is the professionally curated solution for Problem 21 of the 1994 AMC 8, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1994 AMC 8 solutions, or check the answer key.

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Concepts:pigeonhole principle

Difficulty rating: 980

21.

A gumball machine contains 99 red, 77 white, and 88 blue gumballs. The least number of gumballs a person must buy to be sure of getting four gumballs of the same color is

88

99

1010

1212

1818

Solution:

In the worst case, a person could draw 33 red, 33 white, and 33 blue, which is 99 gumballs, without yet having four of any color.

The next (tenth) gumball must match one of these colors, giving four of that color. So 1010 gumballs are needed.

Thus, the correct answer is C .

Problem 21 in Other Years

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