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# Difference between revisions of "2019 AMC 10A Problems/Problem 4"

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

By choosing the maximum number of balls while getting $<15$ of each color, we could have chosen $14$ red balls, $14$ green balls, $14$ yellow balls, $13$ blue balls, $11$ white balls, and $9$ black balls, for a total of $75$ balls. Picking one more ball guarantees that we will get $15$ balls of a color -- either red, green, or yellow. Thus the answer is $75 + 1 = \boxed{\textbf{(B) } 76}$.

## Video Solution 1

Education, The Study of Everything

~savannahsolver