Could the sum of an even number of distinct odd numbers be divisible by each of the odd numbers ?

Let $k\geq 4$ be an even number. Can one find $k$ distinct positive odd numbers $x_1,\ldots,x_k$ such that each $x_i$ divides $S = \sum_{i=1}^k x_i$ ?

Is it possible at least for $k$ big enough ?

  • $\begingroup$ Incidentally this has a solution for k = 0 if you hadn't excluded it with k≥4 $\endgroup$
    – Joshua
    Nov 1, 2017 at 1:30
  • $\begingroup$ @Joshua guess why I excluded it, then... $\endgroup$
    – Olórin
    Nov 1, 2017 at 8:23

1 Answer 1


Yes, it is possible. Divide your sum by $S$ and you have

$$1=\sum_i \frac {x_i}S$$

which is an Egyptian fraction expression of $1$ where all the denominators have the same number of factors of $2$. This is known to be solvable with all denominators odd, but all known solutions have an odd number of terms. A survey paper is here. The section of interest is $9.5$. One example is:

$$1=\frac 13+\frac 15+\frac 17 + \frac 19+\frac 1{11}+\frac 1{15}+\frac 1{35}+\frac 1{45}+\frac 1{231}$$

where the denominators have least common multiple $3465$ so we can write:


with every term dividing the sum. Now if we add $3465$ to each side we have a solution with an even number of terms:


Any Egyptian fraction decomposition of $1$ into fractions with odd denominators yields a solution to your problem. The sum will be twice the least common multiple of the denominators in the decomposition. The paper shows that there is such a decomposition for all odd numbers of terms $9$ or above. You can multiply any solution by any odd number to get another.

What is happening is we are converting the Egyptian fraction decomposition of $1$ with all denominators odd into one that looks like $$1=\frac 12+\frac12\left(\text{all other terms}\right)$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.