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In my previous Question on sums of products of divisors, I asked the following question:

Does it exist some positive integer $n$ such that a sum of $k$ products of its divisors greater than $1$ equal $n$ itself, when $k$ is not a divisor of $n$ itself?

An example was provided, but one of the products was less than $\sqrt{n}$. Subsequently, it came to me the following subsequent question:

Does it exist some positive integer $n$ such that a sum of $k$ products of its divisors greater than $1$ equal $n$ itself, when $k$ is not a divisor of $n$ itself and all the products are equal or greater than $\sqrt{n}$?

Any positive or negative answer to the question, with a counterexample, hint or sketch of a proof in the affirmative or negative direction would be really welcomed. Thanks!

Edit

The answer to the question is yes, as perfectly showed in the answer provided. Would it change if we put the restriction that all the products must be distinct?

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    $\begingroup$ I think the following works :$$2\cdot 3^5\cdot 7^2=2\times ( 3\times 7^2)+\sum_{j=1}^{4}(2\times 7)\times (2\times 3^j\times 7)$$where all the products are distinct, and the smallest term $2\times ( 3\times 7^2)$ is larger than $\sqrt{2\cdot 3^5\cdot 7^2}$. $\endgroup$
    – mathlove
    Dec 20, 2020 at 7:58

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The example you were given scales suitably. That is, where the example had the divisors $(2,3,6,9)$ just multiply them, say, by $5$. We get $(10,15,30,45)$ and the parallel sum is $$10\times 15+10\times 30+10\times 45+15\times 30=150+300+450+450=1350$$

Of course, $1350$ is divisible by each of $(10,15,30,45)$ and now each product is considerably greater than $\sqrt {1350}=36.742\cdots$. And, of course, $4$ still doesn't divide $1350$.

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  • $\begingroup$ thanks for your answer! $\endgroup$ Dec 20, 2020 at 7:20
  • $\begingroup$ Now I wonder if the answer would change if we put the restriction of all the products being distinct $\endgroup$ Dec 20, 2020 at 7:29
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    $\begingroup$ @JuanMoreno There's no difficulty there, just multiply three of the factors by $5$ instead of all four. Thus take $(10,15,30,9)$ and look at the sum $150+300+90+450=990$. $\endgroup$
    – lulu
    Dec 20, 2020 at 11:38

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