Fool's problem of the day ( on request of asnaseer), probably easy , I haven't thought much,

Let $A = \{a_1, a_2 \cdots a_k\} $ be any set of $ k $ composite numbers such that
$1 \le a_i \le 120 $ for all $i$ such that $1 \le i \le k$. Find the least value of $k$ such
that there exists at least one pair $ (a_i, a_j) $, $1\le i, j \le k$ in $A$ which is not co
prime ?

Question

Fool's problem of the day ( on request of asnaseer), probably easy , I haven't thought much,

Let $A =  \{a_1, a_2 \cdots a_k\} $ be any set of $ k $ composite numbers such that
$1 \le a_i \le 120 $ for all $i$ such that $1 \le i \le k$. Find the least value of $k$ such
that there exists at least one pair $ (a_i, a_j) $, $1\le  i, j \le  k$ in $A$ which is not co
prime ?

anonymous · Answer

I should go to bed now, Enjoy guys :)

asnaseer · Answer

good night FFM

asnaseer · Answer

and thanks for the post

anonymous · Answer

You are always  welcome :)

mr.math · Answer

I guess k=4.

asnaseer · Answer

isn't it k=2 --> {1,2}
are 1 and 2 considered to be co prime?

mr.math · Answer

2 is not composite.

asnaseer · Answer

ah! - of course - thx for pointing that out Mr.Math

mr.math · Answer

Neither is 1.

asnaseer · Answer

I thought 1 is coprime?

asnaseer · Answer

but not composite

mr.math · Answer

Two integers a and b are said to be coprime (also spelled co-prime) or relatively prime if the only positive integer that evenly divides both of them is 1.

mr.math · Answer

What about $A=\{4,6\}?$, that's k=2.

asnaseer · Answer

but 2 divides 4 and 6?

mr.math · Answer

Yep, so they are NOT coprime.

mr.math · Answer

As the question asked.

asnaseer · Answer

note to  self: must learn to read the question properly!

asnaseer · Answer

then your answer seems to be correct.

mr.math · Answer

I think so, but that was too easy.
Congrats on being a moderator, they couldn't have chosen any better :-)

asnaseer · Answer

thx Mr.Math - I am very humbled to have been chosen.
FFM did say this is quite easy, but you are right - it seems TOO easy :-)

mr.math · Answer

Plus, he always says that even when it's TOO difficult. So I never trust his judgement. :D

asnaseer · Answer

he he - yes - I concur. I'll have to mull over this one tomorrow. it's quite late here so I need to get some sleep. bye for now...

mr.math · Answer

Good night.

anonymous · Answer

Oh well not that easy ;) Does $ K=2 $ or$ K=4 $ works for any co-prime in$ 1\le a_i \le 120 $ ?

anonymous · Answer

Whenever I say that the problem is easy, it means that there exists a very short solution for that problem which may or may not use some well known theorems or results, and if I remember correctly I have posted only one too difficult problem here.

anonymous · Answer

and now I have solved this one, I would say this is a easy problem ;)