AMC 10 · 2019 · #4

Grade 3 arithmetic

Archetype Small-Domain Constrained Search └ Sub-archetype Sum-Target Search

extremal-constructionmulti-digit-arithmeticsystematic-enumeration identify-subproblemssystematic-enumeration ↑ Prerequisites: multi-digit-arithmetic

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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?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2019 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

Try it yourself first — the explanation is most useful after you’ve attempted it.

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Toolkit + CCSS Solution

Understand

Restated: A box has $28$ red, $20$ green, $19$ yellow, $13$ blue, $11$ white, and $9$ black balls. Drawing without looking and without putting balls back, how many do we need to draw to be sure that some color has been drawn at least $15$ times?

Givens: Color counts: red $= 28$, green $= 20$, yellow $= 19$, blue $= 13$, white $= 11$, black $= 9$; Total balls in the box: $28 + 20 + 19 + 13 + 11 + 9 = 100$; Draws are without replacement; Goal: guarantee at least $15$ balls of one single color; Answer choices: (A) $75$, (B) $76$, (C) $79$, (D) $84$, (E) $91$

Unknowns: The smallest number of draws that forces $15$ of one color

Understand

Plan

Primary tool: #16 Change Focus / Count the Complement

Secondary: #2 Make a Systematic List, #3 Eliminate Possibilities

Tool #16 (Change Focus): instead of asking 'how many draws guarantees $15$ of one color', flip the question — 'what is the largest number of draws where we still avoid having $15$ of any color?' Add one more to that worst case and the next ball MUST push some color to $15$. Tool #2 (List): write out the maximum we can take of each color without hitting $15$ — capped at $14$ for the three big colors, and capped at the whole supply for the three small ones. Tool #3 (Eliminate): the choices $75$, $76$, $79$, $84$, $91$ differ by only a few — once we compute the worst-case total $75$, the answer is $75 + 1 = 76$, picking (B).

Execute — Answer: B

#16 Change Focus / Count the Complement 3.OA.A.3 Step 1

Decide which colors could possibly reach $15$.
Only red ($28$), green ($20$), and yellow ($19$) have at least $15$ balls.
Blue ($13$), white ($11$), and black ($9$) cannot reach $15$ even if we drew every one — they are 'safe' from causing a win.

Capable of $15$: red, green, yellow. Cannot: blue, white, black.

💡 Only colors with $\ge 15$ balls can ever 'cross the line'.

#2 Make a Systematic List 1.OA.A.2 Step 2

Build the worst-case 'avoid-$15$' draw.
For colors that could hit $15$ (red, green, yellow), take the maximum that still stays under $15$ — that's $14$ of each.
For colors that can't hit $15$ anyway (blue, white, black), take ALL of them, since they never cause a win.

Worst case = $\underbrace{14 + 14 + 14}_{\text{red, green, yellow}} + \underbrace{13 + 11 + 9}_{\text{blue, white, black}}$

💡 Pick the most balls you can without any single color reaching $15$.

#2 Make a Systematic List 2.NBT.B.5 Step 3

Add up the worst case.
$14 + 14 + 14 = 42$ for the big three.
$13 + 11 + 9 = 33$ for the small three.
Total $42 + 33 = 75$ balls drawn, and still NO color has reached $15$.

$$14 + 14 + 14 + 13 + 11 + 9 = 42 + 33 = 75$$

💡 $75$ is the largest 'unlucky' draw — still no winning color.

#16 Change Focus / Count the Complement 1.OA.A.1 Step 4

After $75$ draws, only red, green, yellow can still appear (blue/white/black are exhausted).
The $76$th ball is forced to be red, green, or yellow — pushing one of those colors from $14$ to $15$.
So $76$ draws guarantees a color with $15$ balls.

$$75 + 1 = 76 \;\Rightarrow\; \textbf{(B)}$$

💡 One more draw past the worst case forces success.

#3 Eliminate Possibilities 1.NBT.B.3 Step 5

Sanity-eliminate the other choices.
$75$ is the worst-case ceiling for failure, so (A) does NOT guarantee.
$79$, $84$, $91$ all work but are larger than necessary — the question asks for the minimum, so they over-shoot.
Only (B) $76$ is exactly minimum-and-sufficient.

$75$ fails $\Rightarrow$ not (A);\quad $79, 84, 91 > 76$ $\Rightarrow$ not minimum, so not (C), (D), (E)

💡 $76$ is the smallest that always works — the others are too small or too big.

[1] #16 3.OA.A.3 Decide which colors could possibly reach $15$. Only red ($28$), green ($20$), an

[2] #2 1.OA.A.2 Build the worst-case 'avoid-$15$' draw. For colors that could hit $15$ (red, gre

[3] #2 2.NBT.B.5 Add up the worst case. $14 + 14 + 14 = 42$ for the big three. $13 + 11 + 9 = 33$

[4] #16 1.OA.A.1 After $75$ draws, only red, green, yellow can still appear (blue/white/black are

[5] #3 1.NBT.B.3 Sanity-eliminate the other choices. $75$ is the worst-case ceiling for failure,

Review

Reasonableness: Verify the worst-case sum. $14 + 14 + 14 = 42$ and $13 + 11 + 9 = 33$, giving $42 + 33 = 75$ balls with no color at $15$. At draw $76$, only red/green/yellow remain in the box, so one of them goes from $14$ to $15$. The answer $76$ is therefore both achievable (the strategy above attains it) and tight (one more ball is unavoidable).

Alternative: Tool #6 (Guess and Check) directly on the choices. Try (A) $75$: an adversary picks $14$ red, $14$ green, $14$ yellow, and all $13 + 11 + 9 = 33$ of blue/white/black — exactly $75$ balls with no color at $15$, so $75$ fails. Try (B) $76$: now the $33$ 'small' balls are exhausted and the $14$ caps on red/green/yellow cannot all hold — one breaks to $15$. (B) wins.

CCSS standards used (min grade 3)

1.OA.A.1 Solve addition and subtraction word problems within 20 (Adding one more ball to the worst-case total ($75 + 1 = 76$).)
1.OA.A.2 Solve word problems involving three whole numbers whose sum is within 20 (Choosing the maximum-but-still-under-$15$ count for each 'big' color: $14, 14, 14$.)
1.NBT.B.3 Compare two two-digit numbers using symbols (Eliminating the over-shoot choices $79, 84, 91 > 76$.)
2.NBT.B.5 Fluently add and subtract within 100 (Adding the worst-case sum $14 + 14 + 14 + 13 + 11 + 9 = 75$.)
3.OA.A.3 Solve multiplication and division word problems within 100 (Reasoning about the supply caps and which colors can possibly reach $15$.)

⭐ This AMC 10 problem only needs Grade 3 "word-problem reasoning" you already know — count the worst-case 'unlucky' pulls ($14 + 14 + 14 + 13 + 11 + 9 = 75$), then add one more so a color is forced to $15$: $76$.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Review

CCSS standards used (min grade 3)

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