Question 1143356
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There is so called  "the pigeonhole principle"  in Math:


    If 7 pigeons are placed in 6 holes, then at least one hole contains 2 or more pigeons. 


In this joking form it is obvious and does not require more detailed proofs / explanations.




Further, in more general form, 


    if (n+1) pigeons are placed in "n" holes, then there is at least one hole containing 2 or more pigeons. 


In this form it is obvious, again, and does not require more detailed proofs / explanations.




Now let me formulate even more general <U>THEOREM</U>.


    If there are  n*m+1  items in  "m" containers, then there is at least one container containing  "n+1"  or more  items.



The proof is in three lines.


    If there is <U>NO</U> such a container, then the total number of items in "m" containers is <U>NOT MORE</U> than  n*m.


    It  <U>CONTRADICTS</U>  to the <U>given</U> part (!)


    This contradiction proves the statement.




Now let's return to our problem.


We have 731 = 2*365 + 1  students  (= items),  and

        365 days in the year (that are "containers" in this case).


From the <U>Theorem</U>, there is at least one container containing (2+1) = 3 or more items.

In other words, there is at least one day (one date in an year), when 3 or more students celebrate their birthdays.
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Solved.



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In &nbsp;MATHEMATICS &nbsp;(notice all letters are capital, &nbsp;which means that I am talking about &nbsp;TRUE &nbsp;Math)


this principle is called &nbsp;"the Dirichlet's principle", &nbsp;after the famous mathematician &nbsp;Johann &nbsp;Peter &nbsp;Gustav &nbsp;Lejeune &nbsp;Dirichlet 
(1805 - 1859).



About &nbsp;Dirichlet, &nbsp;see these Internet articles

http://www-history.mcs.st-and.ac.uk/Biographies/Dirichlet.html

https://en.wikipedia.org/wiki/Peter_Gustav_Lejeune_Dirichlet



About &nbsp;"the Dirichlet's principle", &nbsp;see this Internet article

https://en.wikipedia.org/wiki/Pigeonhole_principle