2011 AMC 10B Problem 11
There are people in a room. What is the largest value of such that the statement "At least people in this room have birthdays falling in the same month" is always true?
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Key insight
If every month had at most 4 birthdays, the room would hold at most 48 people. With 52, some month has 5, and 6 can fail.
Solution
Think of the months as boxes and put each person in the box for their birthday month. We need two things: show that is always true, and show that is not always true.
Part 1: at least in one month is always true.
- Suppose, just to test it, that every month had people or fewer.
- Then the room would hold at most people.
- But there are people. So some month must have or more.
Part 2: at least in one month can fail.
- Spread the people out as evenly as possible. , so put people in every month, then add one more person to each of four months.
- Now four months have people and eight months have people. That is people.
- No month has people, so "at least " is not always true.
So the largest is .
The answer is .
Why this works
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.
Concise solution
A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.
Think of the months as boxes and the people as objects placed in them.
If every month held at most people, there would be at most people, fewer than . So some month has at least people, no matter how the birthdays fall. The statement is always true for .
Is the largest such ? Distribute people as in four months and in each of the other eight months. Then no month has people, so "at least " can fail.
The answer is .
Review in the AoPS books
- Intermediate Counting & Probability, Chapter 2: Sets and Logic (§2.2, §2.3)
- Introduction to Counting & Probability, Chapter 1: Counting Is Arithmetic
- Intermediate Counting & Probability, Chapter 5: The Pigeonhole Principle
- Prealgebra, Chapter 7: Ratios, Conversions, and Rates (§7.4)
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Problem © Mathematical Association of America (MAA), American Mathematics Competitions. Reproduced for non-commercial educational use. Solution and commentary are original to this site.