AMC 10 Step by Step

2011 AMC 10B Problem 11

P11-15~2 minPrint

There are 5252 people in a room. What is the largest value of nn such that the statement "At least nn 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 1212 months as 1212 boxes and put each person in the box for their birthday month. We need two things: show that 55 is always true, and show that 66 is not always true.

Part 1: at least 55 in one month is always true.

  1. Suppose, just to test it, that every month had 44 people or fewer.
  2. Then the room would hold at most 12×4=4812 \times 4 = 48 people.
  3. But there are 5252 people. So some month must have 55 or more.

Part 2: at least 66 in one month can fail.

  1. Spread the people out as evenly as possible. 52=48+452 = 48 + 4, so put 44 people in every month, then add one more person to each of four months.
  2. Now four months have 55 people and eight months have 44 people. That is 4×5+8×4=20+32=524 \times 5 + 8 \times 4 = 20 + 32 = 52 people.
  3. No month has 66 people, so "at least 66" is not always true.

So the largest nn is 55.

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

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 1212 months as boxes and the 5252 people as objects placed in them.

If every month held at most 44 people, there would be at most 12⋅4=4812 \cdot 4 = 48 people, fewer than 5252. So some month has at least 55 people, no matter how the birthdays fall. The statement is always true for n=5n = 5.

Is 55 the largest such nn? Distribute 52=4⋅12+452 = 4 \cdot 12 + 4 people as 5,5,5,55, 5, 5, 5 in four months and 44 in each of the other eight months. Then no month has 66 people, so "at least 66" can fail.

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

Review in the AoPS books

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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.