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2011 AMC 10B Problem 11

2011 AMC 10B Problem 11AMC 10 Step by Step

Name: Date: September 13, 2026

Sets, Estimation & Miscellaneous · difficulty 2 of 5 · about 2 min · position P11-15

There 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?

A)
$2$
B)
$3$
C)
$4$
D)
$5$
E)
$12$

Answer

Key insight

Twelve months can hold at most 4 people each with 48 people total, so 52 people force some month to have at least 5, and 5 is sharp.

Solution

Think of the $12$ months as boxes and the $52$ people as objects placed in them.

If every month held at most $4$ people, there would be at most $12 \cdot 4 = 48$ people, fewer than $52$ . So some month has at least $5$ people, no matter how the birthdays fall. The statement is always true for $n = 5$ .

Is $5$ the largest such $n$ ? Distribute $52 = 4 \cdot 12 + 4$ people as $5, 5, 5, 5$ in four months and $4$ in each of the other eight months. Then no month has $6$ people, so "at least $6$ " can fail.

The answer is $\boxed{\textbf{(D)}\ 5}$ .

Why this works

This is the pigeonhole principle in its counted form: $N$ objects in $k$ boxes force some box to hold at least $\lceil N/k \rceil$ . A "largest guaranteed $n$ " question always has two halves, a proof that $n$ works (the pigeonhole bound) and a construction showing $n + 1$ can fail (an even-as-possible spread).

The trap

Answering 4 from 52/12 = 4.33 rounded down, or 12 from misreading the question as the number of months.

Common mistakes

Answering $4$ from $52/12 = 4.33$ rounded down, or $12$ from misreading the question as the number of months.
Giving a value like $2$ or $3$ that is true but not the largest guaranteed value.

Techniques

Consider the largest/smallest element or boundary case · Pigeonhole principle