
1. Let $n$ be a natural number whose all positive divisors are denoted as $d_{1}, d_{2}, \ldots, d_{k}$ in such a way that $d_{1}<d_{2}<\cdots<d_{k}$ (thus $d_{1}=1$ and $d_{k}=n$ ). Determine all the values of $n$ for which both equalities $d_{5}-d_{3}=50$ and $11 d_{5}+8 d_{7}=3 n$ hold.

(Matúš Harminc)


Solution. We distinguish whether $n$ is odd or even.

(i) The case of $n$ odd. Since all the $d_{i}$ 's are odd too, it follows from $11 d_{5}+8 d_{7}=3 n$ that $d_{7} \mid 11 d_{5}$ as well as $d_{5} \mid 8 d_{7}$, hence $d_{5} \mid d_{7}$. In view of $d_{7}>d_{5}$, the relations $d_{5}\left|d_{7}\right| 11 d_{5}$ imply that $d_{7}=11 d_{5}$. Substituting this into $11 d_{5}+8 d_{7}=3 n$, we obtain $99 d_{5}=3 n$ or $33 d_{5}=n$. Thus the four numbers $1,3,11$ and 33 are divisors of $n$, more exactly all its divisors smaller that 50 , since the fifth divisor $d_{5}$ satisfies $d_{5}=d_{3}+50>50$. Consequently, it holds that $d_{1}=1, d_{2}=3, d_{3}=11, d_{4}=33$, $d_{5}=d_{3}+50=61$, and thus $n=33 d_{5}=33 \cdot 61=2013$. The number 2013 is satisfactory indeed, because its first small divisors as indicated in the last sentence and moreover, the subsequent divisors are $d_{6}=61 \cdot 3$ and $d_{7}=61 \cdot 11$; hence $d_{7}=11 d_{5}$ as required.

(ii) The case of $n$ even. Now the equality $11 d_{5}+8 d_{7}=3 n$ implies that $2 \mid d_{5}$ and hence $2 \mid d_{5}-50=d_{3}$ as well. Since $d_{1}=1, d_{2}=2$ and $d_{3} \neq 3$, we conclude that either $d_{3}=4$, or $d_{3}=2 t$, with some integer $t>2$. But the last is impossible (otherwise $t$ is a divisor of $n$ with $d_{2}y)$ are some positive divisors of $n$ again. Substituting this into $11 d_{5}+8 d_{7}=3 n$, we obtain (after cancelling $n$ ) an equation $11 / x+8 / y=3$ which can be solved in a standard way, for example by a simple factorization:

$$
8 x=y(3 x-11) \Leftrightarrow 8(3 x-11)+88=3 y(3 x-11) \Leftrightarrow(3 x-11)(3 y-8)=88 .
$$

The first equation implies that $3 x-11>0$ and hence $3 y-8>0$ as well. Note that $x \geqslant y+1$ yields $3 x-11 \geqslant 3 y-8>0$. Taking in account the prime factorization $88=2^{3} \cdot 11$, we conclude that the ordered pair $(3 x-11,3 y-8)$ of factors must belong to the following set

$$
\{(88,1),(44,2),(22,4),(11,8)\} \text {. }
$$

However, congruences modulo 3 imply the only two pairs $(88,1)$ and $(22,4)$ are admissible. The corresponding pairs $(x, y)$ are $(33,3)$ and $(11,4)$, respectively.

If $(x, y)=(33,3)$, then $d_{5}=n / 33$ (and $\left.d_{7}=n / 3\right)$, thus $1,3,11$ and 33 of divisors of $n$ which leads (as in the above solution) to the solution $n=2013$.

If $(x, y)=(11,4)$, then $d_{5}=n / 11$ and $d_{7}=n / 4$, thus $1,2,4,11,22$ and 44 are divisors of $n$, which contradicts to $d_{5}>50$.
