AMC 10 · 2011 · #11
Grade 6 arithmeticThere are 52 people in a room. what is the largest value of n such that the statement "At least n people in this room have birthdays falling in the same month" is always true?
Pick an answer.
AMC 10 2011 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: In a room of $52$ people, every birthday lands in one of the $12$ months. Find the largest number $n$ for which the sentence "at least $n$ people share the same birth month" is guaranteed to be true, no matter how the birthdays happen to fall.
Givens: There are $52$ people in the room.; Every birthday falls in one of the $12$ months of the year.; The claim must hold for every possible arrangement of birthdays, since "always true" allows no exceptions.; Answer choices: (A) $2$, (B) $3$, (C) $4$, (D) $5$, (E) $12$.
Unknowns: The largest $n$ such that some month is guaranteed to contain at least $n$ people.
Understand
Restated: In a room of $52$ people, every birthday lands in one of the $12$ months. Find the largest number $n$ for which the sentence "at least $n$ people share the same birth month" is guaranteed to be true, no matter how the birthdays happen to fall.
Givens: There are $52$ people in the room.; Every birthday falls in one of the $12$ months of the year.; The claim must hold for every possible arrangement of birthdays, since "always true" allows no exceptions.; Answer choices: (A) $2$, (B) $3$, (C) $4$, (D) $5$, (E) $12$.
Plan
Primary tool: #14 Extreme Principle
Secondary: #3 Eliminate Possibilities
"Guaranteed no matter what" is a signal for Tool #14 (Extreme Principle): imagine the arrangement that tries hardest to keep every month small, and see what the fullest month is forced to reach even then. The fairest split is found by dividing the people as evenly as possible across the months. Tool #3 (Eliminate Possibilities) closes the case by showing that a valid arrangement exists whose fullest month is exactly this number, so nothing larger can be guaranteed.
Execute — Answer: D
6.EE.B.5 Step 1 Aim for the fairest split
- To find what is guaranteed, picture the arrangement that fights the statement hardest: spread the $52$ birthdays as evenly as possible over the $12$ months.
- If even this most-balanced arrangement forces some month to hold $n$ people, then every arrangement does, because any lopsided one only piles more into a single month.
💡 A claim that must always hold is decided by the stingiest arrangement, so build that one first.
4.NBT.B.6 Step 2 Divide people among months
- Share the people out evenly by dividing $52$ by $12$.
- That gives $4$ with a remainder of $4$: twelve months each take $4$ people, using $12 \times 4 = 48$, and $4$ people are still left over.
💡 Dividing tells you the even baseline; the remainder is the part that cannot be shared out flatly.
4.OA.A.3 Step 3 Place the leftovers
- The $4$ leftover people still need a month, and each one raises some month from $4$ up to $5$.
- Even in this fairest split, four of the months end up holding $5$ people.
- So no matter how the birthdays fall, some month is forced to have at least $5$ people.
💡 Leftovers cannot vanish; each one pushes a month past the even baseline.
3.OA.A.3 Step 4 Check that six is not forced
- Could we guarantee $6$?
- No.
- The split of eight months with $4$ people and four months with $5$ people already accounts for everyone, since $8 \times 4 + 4 \times 5 = 32 + 20 = 52$, and its fullest month holds only $5$.
- Because a valid arrangement exists whose biggest month is $5$, the number $6$ is not always reachable.
- So the largest value that is always true is $5$, which is choice (D).
💡 If one legal arrangement tops out at $5$, then $6$ can be dodged, so $5$ is the most you can promise.
6.EE.B.5 To find what is guaranteed, picture the arrangement that fights the statement ha 4.NBT.B.6 Share the people out evenly by dividing $52$ by $12$. That gives $4$ with a rema 4.OA.A.3 The $4$ leftover people still need a month, and each one raises some month from 3.OA.A.3 Could we guarantee $6$? No. The split of eight months with $4$ people and four m Review
Reasonableness: The average number of people per month is $52 \div 12 \approx 4.3$, so the busiest month must be at least the next whole number up, which is $5$. That matches the answer. The choices bracket it sensibly: $2$, $3$, $4$ are too weak to be the largest guarantee, and $12$ would need $52$ people to split into at least $12$ per month, which is impossible.
Alternative: Use the Pigeonhole Principle directly: with $52$ pigeons in $12$ holes, some hole holds at least $\left\lceil \dfrac{52}{12} \right\rceil = \lceil 4.33\ldots \rceil = 5$ pigeons. The ceiling of the quotient gives the guaranteed minimum in one step, again yielding $5$, choice (D).
CCSS standards used (min grade 6)
6.EE.B.5Understand solving an equation or inequality as a process of finding values that make it true (Reading "the largest $n$ that is always true" as a boundary condition and testing which value of $n$ the worst-case arrangement still satisfies.)4.NBT.B.6Find whole-number quotients and remainders with up to four-digit dividends (Computing $52 \div 12 = 4$ remainder $4$ to find the even baseline and the leftovers.)4.OA.A.3Solve multi-step word problems using four operations with whole numbers (Reasoning that the $4$ leftover people each push a month from $4$ up to $5$, forcing a fullest month of $5$.)3.OA.A.3Solve multiplication and division word problems within 100 (Checking that $8 \times 4 + 4 \times 5 = 52$ so an arrangement with maximum $5$ really exists, ruling out $6$.)
⭐ Spread things as evenly as you can; the leftovers force the fullest bin up, so $52$ people over $12$ months always crowd $5$ into one month.
⭐ Spread things as evenly as you can; the leftovers force the fullest bin up, so $52$ people over $12$ months always crowd $5$ into one month.
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