Problem 1036

If there exists a positive integer $p \leq\lfloor n / 6\rfloor$ such that $p \nmid n$, then we have $\lfloor n / 2\rfloor>\lfloor n / 6\rfloor$, and taking $k=\lfloor n / 2\rfloor-p \geq 2$ and two positive divisors $d, d+k$ of $n$, we need $d+(\lfloor n / 2\rfloor-p)$ to divide $n$. But $d+(\lfloor n / 2\rfloor-p) \geq d+\lfloor n / 2\rfloor-\lfloor n / 6\rfloor>d+(n / 2-1)-n / 6 \geq n / 3$, so $d+(\lfloor n / 2\rfloor-p) \in\{n / 2, n\}$, the only possible divisors of $n$ larger than $n / 3$. However, $d+(\lfloor n / 2\rfloor-p)=n / 2$ yields $d=p$, absurd (since $d \mid n$ but $p \nmid n$), while $d+(\lfloor n / 2\rfloor-p)=n$ yields $d>n / 2$, thus $d=n$ (since $d \mid n$), forcing $p=\lfloor n / 2\rfloor>\lfloor n / 6\rfloor$, again absurd. Therefore all positive integers not larger than $\lfloor n / 6\rfloor$ must divide $n$.

Denote $u=\lfloor n / 6\rfloor$. Since $\operatorname{gcd}(u, u-1)=1$, it follows $u(u-1) \mid n$, so $u(u-1) \leq n=6(n / 6)1$ cannot be odd (which is trivial), and by the fact that the only $k$ defect for $n=36$ is $k=13$, the idea for the solution given above. Anyway, a fresh idea, and which is implemented elegantly and with minimal calculations.