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We want to choose $n$ number from the four digit numbers. What's the minimum of $n$ if we make sure that at least there are 5 numbers with two same digit in the end ? (for example : $1011$ , $1022$)

My try : I really don't have any idea!

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    Have you ever heard of the pigeon hole principle? This is a direct application of that. What are the pigeons in this case? What are the holes? How many holes are there? How many pigeons then must we use to ensure at least 5 pigeons in at least one of the holes? – 2017-01-25
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    @JorgeFernándezHidalgo Yes , that's right – 2017-01-25
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    @JMoravitz I know that it's related to pigeon hole principle but I can't use it for solving this problem. – 2017-01-25
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    Why can't you? You aren't allowed? Or you simply don't understand how to use it? Look back at my initial comment and answer **every** leading question that I gave. – 2017-01-25
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    It's improper for me to using this principle for solving questions in general but I'll use your questions – 2017-01-25
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    "Its improper..." Who ever gave you that idea? The pigeon-hole principle is one of the more powerful tools that we have to solve questions in combinatorics and is perfectly valid to use in practically any situation that could call for it. I *vehemently* disagree with the absurd notion that the pigeonhole principle could ever be considered "improper for use." – 2017-01-25
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    I know that is powerful tool but I said "for me". Because I can solve equations better than problems like this. – 2017-01-25

2 Answers 2

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There are two possible interpretations I can think of for this question. The first:

How many four-digit numbers must we pick in order to ensure we have at least five numbers $a,b,c,d,e$ chosen with repeating final digits, that is to say the ones digit of $a$ is the same as the tens digit of $a$. The ones digit of $b$ is the same as the tens digit of $b$, but the ones digit of $a$ is not necessarily the same as the ones digit of $b$. E.g. the five numbers could be $\{1011,1022,1033,1044,1055\}$

Here, clearly we could pick all of the numbers without repeating final digits at which point by picking any five additional numbers of the remaining they will all necessarily have repeating final digits since those are the only numbers left to choose from.

That this provides a minimum should be obvious since we may select all of the numbers without final repeating digits and only four of the numbers with final repeating digits. That this amount always forces us to have five numbers with final repeating digits is also clear since if we suppose we have less than five with final repeating digits, that would imply that we have chosen more four digit numbers without final repeating digits than actually exist.

As there are $9\cdot 10\cdot 10\cdot 9=8100$ four digit numbers without repeated final digits, by picking $8105$ four digit numbers, we are guaranteed at least five of them will have repeated final digits.

How many four-digit numbers must we pick in order to ensure we have at least five numbers $a,b,c,d,e$ chosen where they are all equal modulo 100. That is to say, the tens digit of $a$ is the same as the tens digit of $b$ and the ones digit of $a$ is the same as the ones digit of $b$ but the ones digit of $a$ is not necessarily the same as the tens digit of $a$... E.g. the five numbers could be $\{1052,2352,5552, 8052, 9752\}$

In this case, approach via the strong form of the pigeon-hole principle. The pigeons in this case are the numbers we choose. The holes in this case are the possible final two digits. There are $100$ holes.

The strong form of the pigeon-hole principle states that with $p$ pigeons and $h$ holes there must be at least $\lceil \frac{p}{h}\rceil$ pigeons in at least one of the holes. It goes on to say that in order to have at least $n$ pigeons in at least one hole, the minimum number of pigeons required to force that will be $(n-1)\cdot h+1$.

As there are $100$ holes and we wish to have a hole with at least five pigeons in it, we need a minimum of $(5-1)\cdot 100 + 1=401$ pigeons to guarantee that will happen.

That is to say, we need to pick at least $401$ four digit numbers to ensure that at least five of them have the same final two digits as one another.

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The worst case scenario is getting every four-digit number that doesn't have the last two digits equal. Then, just add $5$. I assume you can calculate how many 4-digit numbers do not have third and fourth digits that match?