AMC 10 · 2010 · #3
Grade 1 countingA drawer contains red, green, blue, and white socks with at least 2 of each color. What is
the minimum number of socks that must be pulled from the drawer to guarantee a matching
pair?
Pick an answer.
AMC 10 2010 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 drawer holds socks in 4 colors — red, green, blue, and white — with at least 2 socks of every color. Pulling socks out one at a time without looking, find the smallest number of pulls that makes a matching pair certain, no matter how unlucky the order is.
Givens: There are exactly 4 colors: red, green, blue, and white; The drawer has at least 2 socks of each color, so a matching pair of any color is possible; Socks are pulled blindly, so you cannot choose which color comes next; Answer choices: (A) $3$, (B) $4$, (C) $5$, (D) $8$, (E) $9$
Unknowns: The minimum number of socks that must be pulled to guarantee two of the same color
Understand
Restated: A drawer holds socks in 4 colors — red, green, blue, and white — with at least 2 socks of every color. Pulling socks out one at a time without looking, find the smallest number of pulls that makes a matching pair certain, no matter how unlucky the order is.
Givens: There are exactly 4 colors: red, green, blue, and white; The drawer has at least 2 socks of each color, so a matching pair of any color is possible; Socks are pulled blindly, so you cannot choose which color comes next; Answer choices: (A) $3$, (B) $4$, (C) $5$, (D) $8$, (E) $9$
Plan
Primary tool: #14 Extreme Principle
Secondary: #3 Eliminate Possibilities
The word "guarantee" points straight at Tool #14 (Extreme Principle): to be certain, plan for the unluckiest possible order and see how long you can avoid a pair. With only 4 colors you can dodge a match for at most 4 pulls, so the very next pull forces a repeat. Tool #3 (Eliminate Possibilities) then checks the answer against the choices, ruling out the too-small numbers (that only *might* pair) and the too-big numbers (that count far past the first guaranteed match).
Execute — Answer: C
1.OA.A.1 Step 1 Imagine the unluckiest draw
- To be sure of a pair, plan for the worst possible luck: every sock you pull is a brand-new color.
- There are only 4 colors, so the longest you can keep drawing all-different socks is one red, one green, one blue, and one white — that is 4 socks with no pair yet.
💡 You can only stay pair-free while every sock brings a new color, and there are just 4 colors to hand out.
1.OA.C.6 Step 2 Pull one more sock
- After those 4 socks, every color has already appeared once.
- The very next sock — the 5th — cannot bring a new color, because there are no colors left.
- It must repeat a color you already hold, which finishes a matching pair.
- So $4+1 = 5$ pulls always work.
💡 Once each color is used up once, there is nowhere new for the next sock to go except onto a color you already have.
1.OA.A.1 Step 3 Check against the choices
- Pulling only 4 socks might give one of each color and no pair, so 4 (choice B) and 3 (choice A) do not guarantee anything.
- Numbers like 8 or 9 (choices D and E) would work, but they count far past the first sure pair — the question asks for the smallest count.
- The smallest number that always forces a pair is 5, which is choice (C).
💡 The right answer is the exact turning point: one less still risks no pair, and anything more is overkill.
1.OA.A.1 To be sure of a pair, plan for the worst possible luck: every sock you pull is a 1.OA.C.6 After those 4 socks, every color has already appeared once. The very next sock — 1.OA.A.1 Pulling only 4 socks might give one of each color and no pair, so 4 (choice B) a Review
Reasonableness: The answer should be one more than the number of colors, and it is: 4 colors, so 5 socks. Test the neighbors — 4 socks can come out as red, green, blue, white with no match, proving 4 is too few; and by the 5th sock there is no unused color left, proving 5 is always enough. That the answer lands just above the color count, not at the total sock count, is exactly right for a "guarantee" question.
Alternative: Frame it as the pigeonhole principle: treat the 4 colors as 4 boxes and each pulled sock as an item dropped into its color box. As long as no box holds 2 items you have no pair, which caps you at 4 items (one per box). The next item, the 5th, must fall into an already-occupied box, forcing a pair — again giving 5, choice (C).
CCSS standards used (min grade 1)
1.OA.A.1Solve addition and subtraction word problems within 20 (Reasoning about the worst-case draw — counting that 4 colors allow at most 4 all-different socks — and checking that 4 is too few while 8 and 9 overshoot.)1.OA.C.6Add and subtract within 20 using strategies (Computing $4 + 1 = 5$ to see that one sock beyond the 4 colors forces a repeat.)
⭐ To be sure of a matching pair, plan for the worst luck: with 4 colors you could pull one of each, so the 5th sock has no new color left and must make a pair.
⭐ To be sure of a matching pair, plan for the worst luck: with 4 colors you could pull one of each, so the 5th sock has no new color left and must make a pair.
More like this
Same archetype — closest grade level first.