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Consider positive integers $a_1,a_2,\ldots,a_{1394}$ so that neither of two numbers of $\dfrac{a_1}{a_2},\dfrac{a_2}{a_3},\cdots,\dfrac{a_{1393}}{a_{1394}}$ be equal to each other. at least how many different numbers are there from $a_1,a_2,\cdots,a_{1394}$?

$$1)38\quad\quad\quad\quad\quad\quad 2)45\quad\quad\quad\quad\quad\quad3)49\quad\quad\quad\quad\quad\quad4)53\quad\quad\quad\quad\quad\quad5)55$$

To solve this problem in order to have unique $\dfrac{a_m}{a_{m+1}}$ I considered $a_m$ be prime numbers. so I started form $2$ and wrote few terms of the sequence: $2,3,2,5,3,5,2$ from here I should use new number ($7$). $2,3,2,5,3,5,2,7,3,7,5,7,2$ and here should continue with $11$. I don't know if writing numbers help or not.

Amirali
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1 Answers1

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You are on the right track.

What you need is to realize that essentially you are looking for the smallest $n$ such that $2{n\choose 2}\geq 1394$.

See, the first sequence $2,3,2$ has length of pairs $2=2{2\choose 2}$.

The second sequence $2,3,2,5,3,5,2$ has length of pairs $6=2{3\choose 2}$

The third sequence $2,3,2,5,3,5,2,7,3,7,5,7,2$ has length of pairs $12=2{4\choose 2}$

The result is pretty obvious because you use every pair of numbers exactly twice, once the bigger one first and once the smaller one first.

Thus $2{38\choose 2}=1406$ is the smallest $n$.

cr001
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  • Small detail: I think the formula should have a +1, because you're allowed, if I'm not mistaken, to have the sequence $2,2,3,2$ for the first one, for example. It doesn't affect the answer in this case though. – AnilCh Feb 15 '21 at 10:29