2019 AMC 10A Problems/Problem 4

Revision as of 16:57, 9 February 2019 by Eric2020 (talk | contribs) (Solution)
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 $28$ red balls, $20$ green balls, $19$ yellow balls, $13$ blue balls, $11$ white balls, and $9$ black balls. What is the minimum number of balls that must be drawn from the box without replacement to guarantee that at least $15$ balls of a single color will be drawn$?$

$\textbf{(A) } 75 \qquad\textbf{(B) } 76 \qquad\textbf{(C) } 79 \qquad\textbf{(D) } 84 \qquad\textbf{(E) } 91$

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 $-n$ to $n$ is $0$, so we can choose $n$ to be 44. Then the only number to add left is $45$. Therefore we have:

\[\begin{center} 44+44+1+1 = \boxed{90} \end{center}\] (Error compiling LaTeX. Unknown error_msg)

See Also

2019 AMC 10A (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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

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