# 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

We try to find the worst case scenario where we can find the maximum number of balls that can be drawn while getting $<15$ of each color by applying the pigeonhole principle and through this we get a perfect guarantee. Namely, we can draw up to $14$ red balls, $14$ green balls, $14$ yellow balls, $13$ blue balls, $11$ white balls, and $9$ black balls, for a total of $75$ balls, without drawing $15$ balls of any one color. Drawing one more ball guarantees that we will get $15$ balls of one color — either red, green, or yellow. Thus, the answer is $75 + 1 = \boxed{\textbf{(B) } 76}$.

## Video Solution 1

Education, The Study of Everything

~ pi_is_3.14

## Video Solution 3

~savannahsolver

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 