2015 Azerbaijan IMO TST

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Source: 2015 Azerbaijan IMO TST

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IstekOlympiadTeam

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IstekOlympiadTeam

#1 h May 29, 2015, 11:42 AM • 2 Y

Y by Adventure10, Mango247

We say that $A$ $=$ { $a_1,a_2,a_3\cdots a_n$ } consisting $n>2$ distinct positive integers is $good$ if for every $i=1,2,3\cdots n$ the number ${a_i}^{2015}$ is divisible by the product of all numbers in $A$ except $a_i$ . Find all integers $n>2$ such that exists a $good$ set consisting of $n$ positive integers.

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andria

#4 h May 29, 2015, 12:37 PM • 3 Y

Y by samirka259, ayan_mathematics_king, Adventure10

Trivial problem:
My solution :
If $n\ge 2016$ then assume that there are $n$ distinct positive integers that they satisfy the condition. Then let $S$ product of the numbers and assume that $x$ is minimum number among the $n$ numbers then by the condition $S\mid x^{2016}\longrightarrow x^{2016}<S\le x^{2016}$ contradiction so $n\le 2015$ we show by induction that for any $3\le n \le 2015$ the claim is true for $n=3$ consider $6,18,36$ now assume that the claim is true for $n <2015$ so there are $n$ distinct natural numbers $a_1, a_2,\cdots , a_n$ that satisfy the condition. Let $S=a_1a_2\cdots a_n$ now consider the $n+1$ numbers $b_1=S, b_2=Sa_1, \cdots , b_{n+1}=Sa_n$ clearly all of them are distinct and you can easily observe that they satisfy the condition.
DONE

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17 posts

#5 h Feb 14, 2021, 9:19 AM

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andria wrote:

Trivial problem:
My solution :
If $n\ge 2016$ then assume that there are $n$ distinct positive integers that they satisfy the condition. Then let $S$ product of the numbers and assume that $x$ is minimum number among the $n$ numbers then by the condition $S\mid x^{2016}\longrightarrow x^{2016}<S\le x^{2016}$ contradiction so $n\le 2015$ we show by induction that for any $3\le n \le 2015$ the claim is true for $n=3$ consider $6,18,36$ now assume that the claim is true for $n <2015$ so there are $n$ distinct natural numbers $a_1, a_2,\cdots , a_n$ that satisfy the condition. Let $S=a_1a_2\cdots a_n$ now consider the $n+1$ numbers $b_1=S, b_2=Sa_1, \cdots , b_{n+1}=Sa_n$ clearly all of them are distinct and you can easily observe that they satisfy the condition.
DONE

Sorry to bother , but i think you're solution is wrong because you understood the question the wrong way . It says (a_i)^2015|S

This post has been edited 3 times. Last edited by Mathmatin, Feb 14, 2021, 9:21 AM
Reason: Latex

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#6 h Yesterday at 11:03 AM

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---

For $n > 2015$ , there does not exist any good set $A$ .
For example, let us assume $n = 2016$ and consider the smallest element in $A$ , denoted as $a_1$ . Then, we have

$a_1 a_2 a_3 \dots a_{2016} > a_1^{2016}$
However, according to the given condition,

$a_1 a_2 a_3 \dots a_{2016} \mid a_1^{2016}$
Now, we will show that for every $n \in [3, 2015]$ , there always exists a good set $A$ .
Let

$a_1 = p^{b_1}, \quad a_2 = p^{b_2}, \quad a_3 = p^{b_3}, \quad \dots, \quad a_{k} = p^{b_{k}}$
where

$b_1 < b_2 < b_3 < \dots < b_{k}$
From the given condition, we know that

$a_1 a_2 a_3 \dots a_n \mid a_i^{2016}$
which implies that

$b_1 + b_2 + b_3 + \dots + b_k \leq 2016 b_1$
$b_1 + b_2 + b_3 + \dots + b_k \leq 2016 b_2$
$\vdots$
$b_1 + b_2 + \dots + b_k < 2016 b_k$
Thus, for every $k < 2016$ , the condition holds.

**Answer:** For every $n \in [3, 2015]$ , there exists a good set $A$ .

This post has been edited 2 times. Last edited by MuradSafarli, Yesterday at 11:36 AM

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