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!
Best Answer
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 $n=2\cdot 3^p$.