AMC 8 · 2005 · #16

Grade 4 counting

Archetype Small-Domain Constrained Search

optimization-countingsystematic-enumerationmulti-digit-arithmetic optimization-countingcasework ↑ Prerequisites: systematic-enumeration

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Problem

A five-legged Martian has a drawer full of socks, each of which is red, white or blue, and there are at least five socks of each color. The Martian pulls out one sock at a time without looking. How many socks must the Martian remove from the drawer to be certain there will be 5 socks of the same color?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

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Toolkit + CCSS Solution

Understand

Restated: A Martian's drawer holds red, white, and blue socks with at least five of each color. The Martian pulls socks out one at a time, sight unseen. What is the smallest number of socks the Martian must pull to be guaranteed five socks of one color?

Givens: Three colors are possible: red, white, blue; At least $5$ socks of each color are in the drawer; Socks are pulled one at a time without looking; Answer choices: (A) $6$, (B) $9$, (C) $12$, (D) $13$, (E) $15$

Unknowns: The minimum number of socks pulled that guarantees $5$ of the same color

Understand

Plan

Primary tool: #14 Extreme Principle

Secondary: #9 Solve an Easier Problem, #2 Make a Systematic List

"Be certain" is the tell for Tool #14 (Extreme Principle): the answer is whatever the worst-case sequence forces, not what a lucky pull might give. The worst case is the one that postpones reaching $5$ of any one color as long as possible — spread the pulls as evenly across the three colors as we can. Tool #9 (Solve an Easier Problem) makes the idea concrete by warming up on a smaller version (say, $3$ of one color from $2$ colors) so the pattern is visible. Tool #2 (Systematic List) records the worst-case tally color by color so the count is hard to miscount.

Execute — Answer: D

#9 Solve an Easier Problem 3.OA.A.1 Step 1

Warm up on an easier version.
Suppose only $2$ colors and you want $3$ of one color.
The unluckiest pulls go $2$ of each color first — that is $2 \times 2 = 4$ socks with no color yet at $3$.
The very next sock (the $5$th) must be the $3$rd of some color.
So the answer to the easier version is $(\text{target} - 1) \times (\text{colors}) + 1 = 2 \times 2 + 1 = 5$.
That formula will travel.

$$\text{easier case: } (3-1) \times 2 + 1 = 5 \text{ socks}$$

💡 Grade 3 sees multiplication as "equal groups." Here the groups are colors, and each group can hold up to $\text{target} - 1$ socks before the rule kicks in.

#2 Make a Systematic List 3.OA.A.1 Step 2

Build the worst case for the real problem.
With $3$ colors and a target of $5$ of one color, the unluckiest pulls give $4$ of each color before any color reaches $5$.
List the running tallies color by color to see this is the most you can pull without hitting the goal.

$$\text{worst case tally: red } 4,\ \text{white } 4,\ \text{blue } 4 \;\Rightarrow\; 4 + 4 + 4 = 12 \text{ socks}$$

💡 $3$ groups of $4$ is the Grade 3 "equal groups" picture: $3 \times 4 = 12$. None of the three colors has reached $5$ yet.

#14 Extreme Principle 3.OA.D.8 Step 3

Apply the extreme principle: one more sock decides it.
After $12$ socks with $4$ of each color, the next sock pulled (the $13$th) must be red, white, or blue — and whichever it is, that color jumps from $4$ to $5$.
So $13$ socks always guarantee $5$ of one color, and $12$ does not.
The smallest guaranteed count is $13$.

$$12 + 1 = 13 \;\Rightarrow\; \textbf{(D)}$$

💡 Worst case plus one is the Grade 3 two-step move: compute the unluckiest total, then add $1$ to push it past the threshold.

#14 Extreme Principle 4.OA.A.3 Step 4

Confirm that fewer socks are not enough.
With only $12$ socks pulled, the worst-case split $(4,4,4)$ leaves every color stuck at $4$ — no color has reached $5$.
So $12$ cannot be a guarantee, and any answer smaller than $13$ fails for the same reason.

$$12 = 4+4+4 \;\Rightarrow\; \text{no color at } 5 \;\Rightarrow\; 12 \text{ is not a guarantee}$$

💡 Showing a single bad scenario is enough to knock out a candidate answer — the Grade 4 "check whether the answer makes sense" habit, used in reverse.

[1] #9 3.OA.A.1 Warm up on an easier version. Suppose only $2$ colors and you want $3$ of one co

[2] #2 3.OA.A.1 Build the worst case for the real problem. With $3$ colors and a target of $5$ o

[3] #14 3.OA.D.8 Apply the extreme principle: one more sock decides it. After $12$ socks with $4$

[4] #14 4.OA.A.3 Confirm that fewer socks are not enough. With only $12$ socks pulled, the worst-

Review

Reasonableness: Plug into the same formula as the warm-up: $(\text{target} - 1) \times (\text{colors}) + 1 = (5 - 1) \times 3 + 1 = 4 \times 3 + 1 = 13$. The formula matches the step-by-step argument and matches choice (D). Check the nearby choices: (C) $12$ misses by exactly one (the worst case $(4,4,4)$), and (E) $15$ overshoots — by sock $13$ the goal is already guaranteed. The five legs and "at least five of each color" in the problem only rule out the drawer running out, which never happens before $13$ pulls.

Alternative: Tool #16 (Change Perspective): instead of pulling socks, think of dropping each sock into one of three color bins. The question becomes: how many drops force some bin to hold $5$? If every bin holds at most $4$, the total is at most $3 \times 4 = 12$. So $12$ drops can leave every bin at $4$, but $13$ drops cannot — at least one bin must hold $5$. Same answer, $13$, framed as a Grade 3 division-with-remainder picture ("$13$ shared among $3$ bins forces a bin with at least $\lceil 13/3 \rceil = 5$").

CCSS standards used (min grade 4)

3.OA.A.1 Interpret products of whole numbers, e.g., interpret $5 \times 7$ as the total number of objects in $5$ groups of $7$ objects each (Reading $3$ colors $\times$ $4$ socks per color as $3 \times 4 = 12$ in the worst-case tally, the equal-groups picture that pins down the largest "not yet" total.)
3.OA.D.8 Solve two-step word problems using the four operations (Combining the worst-case multiplication $3 \times 4 = 12$ with the "$+1$" tipping sock to reach the guaranteed total $13$.)
4.OA.A.3 Solve multistep word problems posed with whole numbers using the four operations (Stringing together the warm-up case, the worst-case construction, the "plus one" finish, and the check that $12$ socks is not enough — a multi-step word problem ending in the count $13$.)

⭐ "Be certain" problems are solved by the unluckiest run. Spread your pulls as evenly as possible across the categories, then add one more — that is the guarantee.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Review

CCSS standards used (min grade 4)

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