1. ## Pigeon Hole problem!

how do i show that in any set of 6 integers, there must be at least two whose difference is divisible by 5?

thanks.

Kyle

2. Originally Posted by CSkyle
how do i show that in any set of 6 integers, there must be at least two whose difference is divisible by 5?
thanks.
Kyle
Pigeon Hole Principle
Observe that if two numbers leave the same remainder with 5, then their difference is divisible by 5. Since there are only 5 distinct remainders possible(i.e{0,1,2,3,4}), in a set of six numbers, at least 2 numbers must leave the same remainder when divided by 5(pigeon hole principle). So the conclusion follows.

3. Originally Posted by Isomorphism
Pigeon Hole Principle
Observe that if two numbers leave the same remainder with 5, then their difference is divisible by 5. Since there are only 5 distinct remainders possible(i.e{0,1,2,3,4}), in a set of six numbers, at least 2 numbers must leave the same remainder when divided by 5(pigeon hole principle). So the conclusion follows.
Nice solution.

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