Difference between revisions of "2011 AMC 10B Problems/Problem 11"
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==Solution== | ==Solution== | ||
| − | 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, because there are <math>12</math> months in a year. 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> |
== See Also== | == See Also== | ||
{{AMC10 box|year=2011|ab=B|num-b=10|num-a=12}} | {{AMC10 box|year=2011|ab=B|num-b=10|num-a=12}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 22:20, 16 February 2016
Problem
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?
Solution
Pretend you have people you want to place in 
boxes, because there are months in a year. By the Pigeonhole Principle, one box must have at least 
people 
See Also
| 2011 AMC 10B (Problems • Answer Key • Resources) | ||
| 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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