Art of Problem Solving
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2011 AMC 10B Problems/Problem 11: Difference between revisions

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==Problem==
== Problem 11 ==


There are 52 people in a room. The statement " At least n people in the room have birthdays in the same month " is true for n. What is the largest possible value of n?
There are <math>52</math> people in a room. what is the largest value of <math>n</math> such that the statement "At least <math>n</math> people in this room have birthdays falling in the same month" is always true?


(A) 2 (B) 3 (C) 4 (D) 5 (E) 6
<math> \textbf{(A)}\ 2 \qquad\textbf{(B)}\ 3 \qquad\textbf{(C)}\ 4 \qquad\textbf{(D)}\ 5 \qquad\textbf{(E)}\ 12 </math>


==Solution==
==Solution==


By the Pigeon Hole Principle, this the 52/12 rounded up. That is 5 (D).
Pretend you have <math>52</math> people you want to place  in <math>12</math> boxes. By the [[Pigeonhole Principle]], one box must have at least <math>\left\lceil \frac{52}{12} \right\rceil</math> people <math>\longrightarrow \boxed{\textbf{(D)} 5}</math>

Revision as of 19:38, 25 May 2011

Problem 11

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?

$\textbf{(A)}\ 2 \qquad\textbf{(B)}\ 3 \qquad\textbf{(C)}\ 4 \qquad\textbf{(D)}\ 5 \qquad\textbf{(E)}\ 12$

Solution

Pretend you have $52$ people you want to place in $12$ boxes. By the Pigeonhole Principle, one box must have at least $\left\lceil \frac{52}{12} \right\rceil$ people $\longrightarrow \boxed{\textbf{(D)} 5}$