Let us consider some positive integer $n$ with divisors (other than $1$) $d_1, d_2, ..., d_n$. Lately, it came to me 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?

My "gut feeling" is that is not possible, but I have not been able to ellaborate any proof of it. If all products added are equal, it is clear that its sum is not equal to $n$ unless $k$ divides $n$, but I am having trouble when considering distinct products of divisors.

It can be noticed easily that if $n$ is some prime number, a perfect power, or a semiprime, then a sum as the defined is not possible, as either there are no possible products of divisors, or all the products of divisors are equal, or all the possible sums of products have a number of terms $k$ which is a divisor of $n$.

Any hint or a sketch of a proof would be really welcomed. Thanks!

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    $\begingroup$ You have the divisors being $d_1,\cdots , d_n$ and the number itself being $n.$ Do you really mean the number of divisors of the number to be the number itself? If not should use a different variable for the number of divisors than for the number itself. [For instance for a prime $p$ the number of divisors is $p+1$ and not $p.$] $\endgroup$
    – coffeemath
    Dec 17, 2020 at 0:22
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    $\begingroup$ $1+1+2+2+2=8$ ? $\endgroup$
    – Empy2
    Dec 17, 2020 at 1:05
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    $\begingroup$ What does “$k$ products of its divisors” mean? Can you give an example of the calculation you are envisioning for some values of $n$ and $k$ (even though it won’t satisfy equality with $n$)? $\endgroup$ Dec 17, 2020 at 1:30
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    $\begingroup$ However $2 \times 5 \times 5 = 5 \times 5 + 2 \times 5 + 2 \times 5 + 5$, there are $k=4$ terms in the sum, and $4$ does not divide $2 \times 5 \times 5$. $\endgroup$
    – player3236
    Dec 17, 2020 at 8:36
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    $\begingroup$ There are several things unclear to me about the statement as it is. If a divisor appears, say, twice, can I use it in two different sums (or twice in the same sum)? Can I use it three times? You seem to disallow 1 as a divisor, and also any products of only one divisor. How many terms are permitted in the products? $\endgroup$ Dec 19, 2020 at 10:35

1 Answer 1


If I understand your question correctly, the answer is yes.

For $n=54=2\times 3^3$, we can have $$54=2\times 3+2\times 6+2\times 9+3\times 6$$ where $2,3,6$ and $9$ are its divisors larger than $1$.

There are $k=4$ terms in the sum, and $4$ does not divide $54$.

Added :

There are infinitely many such $n$.

Proof :

For $n=2\cdot 3^p$ where $p$ is a prime number larger than $3$, we can have $$2\cdot 3^p=2\times (2\times 3^1)+2\times (2\times 3^2)+\cdots +2\times (2\times 3^{p-1})+2\times 3$$

There are $k=p$ terms in the sum, and $p$ does not divide $n=2\cdot 3^p$.

  • $\begingroup$ as always, your answer is most helpful! I realized that one of your products is less than $\sqrt{n}$, do you think it could be found some example with the restriction of all the products being equal or greater than $\sqrt{n}$? $\endgroup$ Dec 19, 2020 at 23:46
  • $\begingroup$ If you prefer, rather than in this post you could post some contribution here: math.stackexchange.com/questions/3955576/… $\endgroup$ Dec 19, 2020 at 23:58
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    $\begingroup$ @Juan Moreno : I've added a proof that there are infinitely many such $n$. It seems that the question in your comment has been answered. $\endgroup$
    – mathlove
    Dec 20, 2020 at 4:34
  • $\begingroup$ "chatty comment" this one but very good answer ;) $\endgroup$ Dec 20, 2020 at 9:56

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