AMC 8 · 2005 · #16

Grade 4 counting

Archetype Small-Domain Constrained Search └ Sub-archetype Factor / Divisibility Constraint

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

📏 Short solution 💡 2 insights

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Problem

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?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2005 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Extreme Principle

"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.

1STEP 1

Warm up small: 2 colors, want 3 of one. Worst luck gives 2 of each, then the next pull forces a third — pattern (target−1)×(colors)+1.

easier case: (3-1) × 2 + 1 = 5 socks

2STEP 2

Now the real case: 3 colors, target 5. Worst luck is 4 of each color before any hits 5 — tally red 4, white 4, blue 4 = 12.

worst case tally: red 4, white 4, blue 4 → 4 + 4 + 4 = 12 socks

3STEP 3

Extreme principle: after 12 socks (4 each), the next pull is red, white, or blue and lifts that color to 5 — so 13 guarantees it, 12 can't.

12 + 1 = 13 → (D)

4STEP 4

Check nothing smaller works: with only 12 pulled, the split 4+4+4 leaves every color at 4 — none at 5 — so 12 is no guarantee.

12 = 4+4+4 → no color at 5 → 12 is not a guarantee

Answer

Plug into the same formula as the warm-up: (target - 1) × (colors) + 1 = (5 - 1) × 3 + 1 = 4 × 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.

💡Key takeaway

"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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