Pigeon hole principle application average

Question

Arrange 0-9 on the circular table There is a section where the sum of three adjacent numbers must be 14 or more

I know we can solve this using pigeonhole principle

I know that the pigeon hole is the number 10 I want to know why we have to sum 0+1+..9 =45 x 3=135

And divide it by 10 =13.5

Is this the correct way to solve this problem? If yes i want to know why we have to find the average ?

Answer 1

Your answer is correct computationally, but I don't think the reasoning is clear for you. I find the best way to understand the pigeonhole principle is to try and see what happens if the condition fails :

If for all sections of three adjacent numbers, their sum is at most 13, then let us take the sum of all these section and denote it $x$. We know for sure that $x$ is at most $13\times \text{number of sections}$, and a bit of counting shows that there are $10$ sections, so in the end we get $x \leq 130$. Moreover, we also know that each number from $0$ to $9$ is counted thrice in $x$, so $x = 3\times (0+1+...+9) = 135$. In the end, $x$ has to be both at most $130$ and equal to $135$, which is impossible. Hence there has to be at least one section whose sum is more than $13$ (ie, at least $14$)

Answer 2

Yes.

There are 10 possible totals at the circular table. [Each number can be added to its two neighbours.]

If we add up all the available totals at the table, we will include every digit 3 times giving $3\times45=135$. There are 10 totals so at least one must exceed 13.

Answer 3

You jumped a few steps, so the proof is not complete. What does the $13.5$ mean tell us? We want to take a mean somewhere and we want to take sums of $3$ adjacent numbers.

Note that each digit appears exactly three times when gathering consecutive $3$ numbers: it’s either the first, the second or the third number. So if we take each of the $10$ consecutive triples and sum, we’ll have a total of $3\times(0+1+\ldots+9) = \frac{3\times9\times10}2 = 135$. So the average of the sums is $13.5$. But if no sum is $14$ or more, the average is at most $13$, a contradiction. Thus there’s a sum being $14$ or more.

Specifically using the pidgeonhole principle (without the contradiction which is basically the pidgeonholing), the holes are the $3$-sums and the pidgeons are “$1$”s: since $135 = 10\times13 + 5$, there’s at least one hole with $14$ or more pidgeons, i.e., there’s at least one $3$-sum greater or equal to $14$.