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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 ?

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  • $\begingroup$ Incidentally this has a solution for k = 0 if you hadn't excluded it with k≥4 $\endgroup$ – Joshua Nov 1 '17 at 1:30
  • $\begingroup$ @Joshua guess why I excluded it, then... $\endgroup$ – ujsgeyrr1f0d0d0r0h1h0j0j_juj Nov 1 '17 at 8:23
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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:

$$3465=1155+693+495+385+315+231+99+77+15$$

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

$$6930=3465+1155+693+495+385+315+231+99+77+15$$

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.

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    $\begingroup$ @ujsgeyrr1f0d0d0r0h1h0j0j_juj , then a2=a1 & a3 will be even. $\endgroup$ – Prem Oct 31 '17 at 17:02
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    $\begingroup$ @ujsgeyrr1f0d0d0r0h1h0j0j_juj , but , that is already what is in the answer , 3465 is the sum of the remaining odd numbers .... $\endgroup$ – Prem Nov 1 '17 at 12:11

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