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There are $k$ people in a room, each person picks a day of the year to get a free dinner at a fancy restaurant. $k$ is such that there must be at least one group of six people who select the same day. What is the possible value of such $k$ if the year is a leap year $(366$ days)?

  1. $1465$
  2. $1831$
  3. $1830$
  4. $2197$
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1 Answer

5 votes
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By Generalized Pigeonhole Principle,

If $k$ is a positive integer and $\text{N}$ objects are placed into $k$ boxes, then at least one of the boxes will contain $\left \lceil \frac{\text{N}}{k} \right \rceil$ or more objects.

Here, we need to find $\text{N,}$ and $k=366,$ and $\left \lceil \frac{\text{N}}{k} \right \rceil = 6$

Least value of $\text{N}$ that will satisfy this is $1831.$

3 Comments

How A is also answer.
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The least value is 1831. then how option A is also correct, option D must also true. as it is greater than 1831.  @GO Classes 

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The question is NOT asking for least possible value, it is asking for “What is the possible value of such k”.

Answer will be B,D. It has been corrected now.
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Answer:

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