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2019 AMC 10A Problem 4

2019 AMC 10A Problem 4AMC 10 Step by Step

Name: Date: September 13, 2026

Sets, Estimation & Miscellaneous · difficulty 1 of 5 · about 2 min · position P1-10

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 $?$

A)
$75$
B)
$76$
C)
$79$
D)
$84$
E)
$91$

Answer

Key insight

Worst case: draw 14 of each color that has at least 14 balls and every ball of the small colors, 75 in all; the next ball forces a 15th.

Solution

Ask how many balls can be drawn while avoiding $15$ of any one color. Each color can contribute at most $14$ balls, but a color with fewer than $14$ balls can only contribute all of them:

red, green, yellow: $14$ each, total $42$ ;
blue $13$ , white $11$ , black $9$ : total $33$ .

So $42 + 33 = 75$ balls can be drawn with no color reaching $15$ . This is the maximum such draw, so once $76$ balls are drawn, some color must appear at least $15$ times (the only colors with room to grow are red, green and yellow, and one of them gets a $15$ th ball).

The answer is $\boxed{\textbf{(B)}\ 76}$ .

Why this works

"Guarantee" problems are pigeonhole arguments: find the largest draw that fails, then add one. The extremal (worst) case fills every color to its cap, where the cap is the smaller of $14$ and the number of balls of that color. The distractor $84 = 6 \cdot 14$ is what you get by ignoring the second cap.

The trap

Using 14 for every color (6 times 14 = 84) even though blue, white and black have fewer than 14 balls.

Common mistakes

Using 14 for every color (6 times 14 = 84) even though blue, white and black have fewer than 14 balls.
Reporting $75$ , the largest failing draw, instead of $76$ , the first draw that guarantees success.

Techniques

Consider the largest/smallest element or boundary case · Pigeonhole principle