Problem!!!(BOMC-2)

[sourav das]

Solution:

Let $P(a_i,a_j)\Rightarrow max(a_i,a_j)-min(a_i,a_j)$
Notice that we can make $20.19=480$ pairs from $20$ elements.But we have $P(a_i,a_j)=P(a_j,a_i)$.Hence the number of pairs is $\frac {480}{2}=240$.
Now notice that The maximum value of $a_i-a_j$ is $69$ and minimum is $1$.So from Pigeon Hole Principle, there is a $k$ for which we can write $a_i-a_j=k$ for $\left \lceil \frac{240}{69} \right \rceil=4$ different $P(a_i,a_j)$.

$20*19=380$. Adib, take your time, ok.

[*Mahi*]

Solution:

Let $P(a_i,a_j)\Rightarrow max(a_i,a_j)-min(a_i,a_j)$
Notice that we can make $20.19=480$ pairs from $20$ elements.But we have $P(a_i,a_j)=P(a_j,a_i)$.Hence the number of pairs is $\frac {480}{2}=240$.
Now notice that The maximum value of $a_i-a_j$ is $69$ and minimum is $1$.So from Pigeon Hole Principle, there is a $k$ for which we can write $a_i-a_j=k$ for $\left \lceil \frac{240}{69} \right \rceil=4$ different $P(a_i,a_j)$.

As Sourav da said, no hurry, and don't make such silly mistake in your calculations. Do not let it turn to fatal.

[Phlembac Adib Hasan]

Ooppssss edited.Now my calculation also says it is impossible for twenty numbers.

[sourav das]

Actually I'm little bit confused. Do anyone can give me a rigorous proof that it is impossible?

[*Mahi*]

I think $21$ elements is the correct number, as $\binom {21}{2}=210 \approx 3 (70-1)+1$ which gives the four $k$ property.