Introduction to Pigeonhole Principle

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SUMMARY

The discussion centers on the Pigeonhole Principle, specifically its application in solving problems such as the birthday problem and a custom example involving selecting numbers. The Pigeonhole Principle states that if more items are put into fewer containers than there are items, at least one container must contain more than one item. In the provided example, selecting 51 numbers from the range of 1 to 100 guarantees that at least two of those numbers will not share a common prime factor, illustrating the principle effectively.

PREREQUISITES
  • Understanding of the Pigeonhole Principle
  • Basic knowledge of prime numbers and their properties
  • Familiarity with combinatorial logic
  • Ability to formulate mathematical proofs
NEXT STEPS
  • Explore the application of the Pigeonhole Principle in combinatorial proofs
  • Study the birthday problem in detail and its implications in probability theory
  • Learn about prime factorization and its relevance in number theory
  • Investigate other mathematical principles related to combinatorics, such as the Inclusion-Exclusion Principle
USEFUL FOR

Students studying mathematics, educators teaching combinatorial concepts, and anyone interested in understanding the applications of the Pigeonhole Principle in problem-solving.

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Homework Statement


Give a sample problem its solution employing Pigeonhole Principle


Homework Equations


Pigeonhole Principle


The Attempt at a Solution


We have this homework about pigeonhole principle which hasn't been discussed yet, but we need to present an example and present it in class,, I've been searching, I get the holes and pigeons like logic stuff,, If i were to get an example i would use the birthday problem which is easy,, but i don't know if this is relevant,, if there's anyone who could explain or could give any easy example.. this would really help.. also please provide with explanation and solution tnx
 
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Try to do this one:

Say I choose 51 numbers from 1,2,...,100. Then there will be at least 2 numbers who do not have a common prime factor.
 
so pigeons would be the 100 and 51 would be the pigeon box right? or 2?
so how do i equate it? do i use subset?
 
No, the pigeon boxes are somewhat more complicated.

Hint: the numbers k and k+1 do not have common prime divisor.
 

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