Problem 1.
Determine all composite integers $n>1$ that satisfy the following property: if $d_1, d_2, \ldots, d_k$ are all the positive divisors of $n$ with 1=d1<d2< ... <dk=n  , then di divides  di+1+di+2 for every $1 \leqslant i \leqslant k-2$.

proposed by Santiago Rodriguez, Colombia
 
Solution
Observe that we have
$$d_i | d_{i+1}+d_{i+2} , 1 \leq i \leq k-2$$,
then put $i=k-2$ then
$$d_{k-2} | d_{k-1}+d_{k}=d_{k-1}+n$$
but
$$d_{k-2} |n \Longrightarrow d_{k-2} | d_{k-1} .$$
Again putting $i=k-3$ we get
$$d_{k-3} | d_{k-2}+d_{k-1}$$
but on other hand
$$d_2=\frac{n}{d_{k-1}} | \frac{n}{d_{k-2}} =d_3 , d_2|d_3+d_4 \Longrightarrow d_2|d_4.$$
So $$d_{k-3}|d_{k-1} \Longrightarrow d_{k-3}|d_{k-2} .$$
Now using Induction we get
$$d_1|d_2|d_3.......d_k$$.
Hence $n=p^{k-1}$ is the solution.