<s>[INST] Determining $n$ from its divisors Let $d_i$ be the $i^{th}$ smallest divisor of $n$.
Suppose that $n$ has $3$ prime factors and that:
$$n = d_{13}+d_{14}+d_{15}$$
$$d_{15}+1 = (d_5+1)^3$$
Is it possible to uniquely determine $n$?
I tried doing some stuff with writing $n=p^\alpha q^\beta r^\gamma$ and working modulo different things, but couldn't figure it out.
 [/INST] Well, $n<3d_{15}$ and $d_{15}\mid n$, so either $n=d_{15}$ or $n=2d_{15}$. However, in that first case, $d_{14}+d_{13}$ would be zero and that's impossible. Hence,
$$d_{15}=\frac n2.$$
It follows that $n<4d_{14}$ and because $d_{14}\mid n$ and $d_{14}<n/2$, we must have $d_{14}\in\{n/3,n/4\}$. But if $d_{14}=n/4$, then $d_{13}$ must be as well, which is not the case. Therefore, 
$$d_{14}=n/3$$ 
and
$$d_{13}=n-\frac n2-\frac n3=\frac n6.$$
We now have that $d_1=1$, $d_2=2$ and $d_3=3$. So the three prime factors are $2$, $3$ and some other prime, $p$.

Note that
$$1\equiv \frac n2 + 1\equiv (d_5+1)^3\equiv d_5+1\pmod 3,$$
so $3\mid d_5$.
Assume $d_5=6$, then $n=2((6+1)^3-1)=684=2^2\cdot 3^2\cdot 19$. But $684$ has $3\cdot 3\cdot 2=18$ proper divisors, while $n$ has $16$, so $684$ is not a solution and we conclude that $d_5\neq 6$. 
Therefore, $d_5\ge 9$, so $d_4=6$ and $4,5,8\nmid n$.

Since $n$ cannot have divisors larger than $n/2$, we conclude that $n$ has exactly $16$ proper divisors. Write $n=2\cdot3^ap^b$ with $a,b\ge 1$,  then it follows that
$$(a+1)(b+1)=8.$$
So $\{a,b\}=\{1,3\}$.
If $a=3$, then $9\mid n$ and we must have $d_5=9$. Then $n=2((9+1)^3-1)=1998=2\cdot 3^3\cdot 37$, so $n=1998$ has three prime factors and also satisfies the conditions.

Finally, say $a=1$, so $n=6p^3$ for some prime $p>3$. Then $d_5=p$, but $3\mid d_5$, which is a contradicion.

Conclusion: $n=1998$ is the only integer which satisfies both conditions.
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