1. Let $n$ be a positive integer. Denote all its positive divisors as $d_{1}, d_{2}, \ldots, d_{k}$ such that $d_{1}<d_{2}<\ldots<d_{k}$ (thus $d_{1}=1$ and $d_{k}=n$). Determine all values of $n$ for which $d_{5}-d_{3}=50$ and $11 d_{5}+8 d_{7}=3 n$. (Matúš Harminc)
Solution. We distinguish whether the sought $n$ is odd or even.

(i) Let $n$ be odd, then all $d_{i}$ are odd. From the equality $11 d_{5}+8 d_{7}=3 n$ it follows that $d_{7} \mid 11 d_{5}$ and also $d_{5} \mid 8 d_{7}$, or $d_{5} \mid d_{7}$. From $d_{5}\left|d_{7}\right| 11 d_{5}$, considering $d_{7}>d_{5}$, we have $d_{7}=$ $=11 d_{5}$ and substituting into the equality $11 d_{5}+8 d_{7}=3 n$ we get $99 d_{5}=3 n$ or $33 d_{5}=n$. We see that the four numbers $1,3,11$ and 33 are divisors of the number $n$, and indeed the only divisors less than 50, because for the fifth divisor $d_{5}$ according to the problem statement we have $d_{5}=d_{3}+50>50$. Thus, we have $d_{1}=1, d_{2}=3, d_{3}=11, d_{4}=33, d_{5}=d_{3}+50=61$, and therefore $n=33 d_{5}=33 \cdot 61=2013$. This number indeed satisfies the conditions, as its smallest divisors are correctly listed in the previous sentence, and additionally $d_{6}=61 \cdot 3$ and $d_{7}=61 \cdot 11$, so it is indeed satisfied that $d_{7}=11 d_{5}$, and thus all the required conditions.

(ii) Let $n$ be even. From the equality $11 d_{5}+8 d_{7}=3 n$ it follows that $2 \mid d_{5}$, so also $2 \mid d_{5}-50=d_{3}$. Since $d_{1}=1$ and $d_{2}=2$, $d_{3}$ cannot be 3, so it must be either $d_{3}=4$, or $d_{3}=2 t$, where $t>2$. The latter, however, is not possible (the number $t<d_{3}$ would be missing from the list of the smallest divisors of the number $n$), and therefore it must be $d_{3}=4$. Then, however, $d_{5}=d_{3}+50=54$, and thus $3 \mid n$, although 3 is not among the smallest divisors of the number $n$. Therefore, no even $n$ that satisfies the conditions exists.

Answer. The problem has a unique solution $n=2013$.