Difference between revisions of "2011 AMC 10B Problems/Problem 11"

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== Problem 11 ==
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== Problem==
  
 
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?
 
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?
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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>
 
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>
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== See Also==
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{{AMC10 box|year=2011|ab=B|num-b=10|num-a=12}}

Revision as of 16:13, 4 June 2011

Problem

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

See Also

2011 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions
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