AMC 8 · 2005 · #16
Easy mode Grade 4Problem
Imagine a Martian with five legs. He has a drawer full of socks, and every sock is red, white, or blue. There are at least five socks of each color.
The Martian reaches into the drawer and pulls out socks one at a time without looking.
He wants to be sure he has 5 socks of the same color. What is the smallest number of socks he must pull out so that he is guaranteed to have 5 of one color, no matter how the colors come out?
Pick an answer.
AMC 8 2005 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.
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
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) $15$
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
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.
💡 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.
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.
💡 $3$ groups of $4$ is the Grade 3 "equal groups" picture: $3 \times 4 = 12$. None of the three colors has reached $5$ yet.
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 $13$.
💡 Worst case plus one is the Grade 3 two-step move: compute the unluckiest total, then add $1$ to push it past the threshold.
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.
💡 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.
3.OA.A.1 Warm up on an easier version. Suppose only $2$ colors and you want $3$ of one co 3.OA.A.1 Build the worst case for the real problem. With $3$ colors and a target of $5$ o 3.OA.D.8 Apply the extreme principle: one more sock decides it. After $12$ socks with $4$ 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.1Interpret 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.8Solve 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.3Solve 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.
⭐ "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.
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