63rd Czech and Slovak Mathematical Olympiad

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 $11d_5+8d_7=3n$ that $d_7\mid 11d_5$ as well as $d_5\mid 8d_7$, hence $d_5\mid d_7$. In view of $d_7>d_5$, the relations $d_5\mid d_7\mid 11d_5$ imply that $d_7=11d_5$. Substituting this into $11d_5+8d_7=3n$, we obtain $99d_5=3n$ or $33d_5=n$. Thus the four numbers 1, 3, 11 and 33 are divisors of $n$, more exactly all its divisors smaller than 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 $n=33d_5=33\cdot 61=2013$. 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 $d_7=11d_5$ as required.

ii) The case of $n$ even. Now the equality $11d_5+8d_7=3n$ 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 $d_3\ne 3$, we conclude that either $d_3=4$, or $d_3=2t$, with some integer $t>2$. But the last is impossible (otherwise $t$ is a divisor of $n$ with $d_2<t<d_3$, a contradiction), Therefore, we have $d_3=4$, $d_5=d_3+50=54$ and hence 3 is a divisor of $n$ between $d_2$ and $d_3$, a contradiction. In this way, the nonexistence of any satisfactory even $n$ is proven.

Answer. The problem has the only solution $n=2013$.

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Alternative solution.

The divisors $d_5$ and $d_7$ of $n$, with $d_5<d_7$, can be represented as $d_5=n/x$ and $d_7=n/y$, where $x$ and $y$ ($x>y$) are some positive divisors of $n$ again. Substituting this into $11d_5+8d_7=3n$, 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: $$8x=y(3x-11)\iff 8(3x-11)+88=3y(3x-11)\iff (3x-11)(3y-8)=88.$$ The first equation implies that $3x-11>0$ and hence $3y-8>0$ as well. Note that $x\ge y+1$ yields $3x-11\ge 3y-8>0$. Taking into account the prime factorization $88=2^3\cdot 11$, we conclude that the ordered pair $(3x-11,3y-8)$ of factors must belong to the following set $$\{(88,1),(44,2),(22,4),(11,8)\}.$$ 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 $d_7=n/3$), thus 1, 3, 11 and 33 are 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 $d_5>50$.