2011 AMC 10B Problem 11

Think of the $12$ months as $12$ boxes and put each person in the box for their birthday month. We need two things: show that $5$ is always true, and show that $6$ is not always true.

Part 1: at least $5$ in one month is always true.

1. Suppose, just to test it, that every month had $4$ people or fewer.
2. Then the room would hold at most $12 \times 4 = 48$ people.
3. But there are $52$ people. So some month must have $5$ or more.

Part 2: at least $6$ in one month can fail.

1. Spread the people out as evenly as possible. $52 = 48 + 4$, so put $4$ people in every month, then add one more person to each of four months.
2. Now four months have $5$ people and eight months have $4$ people. That is $4 \times 5 + 8 \times 4 = 20 + 32 = 52$ people.
3. No month has $6$ people, so "at least $6$" is not always true.

So the largest $n$ is $5$.

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

This is the pigeonhole idea: if you put more things into boxes than the boxes can hold at a certain level, some box must go over that level. For "the largest number that is always true," you need both halves: prove the number always works, then show an example where one more fails.

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 $\boxed{\textbf{(D)}\ 5}$.