3
$\begingroup$

Fix a $k \in \mathbb{N}$

  • How do I find all sets of positive integers $a_{1},a_{2},...,a_{k}$ such that the sum of any triplet is divisible by each member of the triplet.

I couldn't see any pattern. But the simplest example which I could see is the set $\{1,2,3\}$. This satisfies the hypothesis. And even sets like $\{2,4,6\}$ does satisfy.

In search of finding a solution I attempted the following: Let $\{a_{1},a_{2},a_{3}\}$ be a triplet. Then by our hypothesis we should have $$a_{1}\mid (a_{1}+a_{2}+a_{3}), \quad a_{2} \mid (a_{1}+a_{2}+a_{3}) \quad a_{3}\mid (a_{1}+a_{2}+a_{3})$$ Using all this I arrive at $$k_{1}a_{1} = k_{2}a_{2} = k_{3}a_{3}$$ But all this doesn't seem to help.

$\endgroup$
  • $\begingroup$ You write "sets" but don't use set notation in "$a_1,a_2,\ldots,a_k$". Are the $a_i$ meant to be distinct? $\endgroup$ – joriki Jun 2 '16 at 15:06
  • $\begingroup$ @joriki I have basically copied the question from Ivan Niven's Number theory book. And he doesn't use the set notation there. But let's assume $a_{i}$'s are distinct. This is Section 1.3 Exercise 46, from the book :) $\endgroup$ – crskhr Jun 2 '16 at 15:08
6
$\begingroup$

If $a_1\lt a_2\lt a_3$, their sum $S$ satisfies $a_3\lt S\lt 3a_3$. Since $a_3\mid S$, this implies $S=2a_3$, thus $a_1+a_2=a_3$ and thus $a_2\gt a_3/2$. The only divisor of $2a_3$ greater than $a_3/2$ and less than $a_3$ is $\frac23a_3$. Thus we have $a_2=2a_1$ and $a_3=3a_1$. This cannot continute to hold if we add a fourth number, so the only sets with this property are those of the form $\{a_1,2a_1,3a_1\}$ with $a_1\in\mathbb N$.

$\endgroup$

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.