A number $n$ has $12$ divisors and $d_{d_4-1} = (d_1+d_2+d_4)d_8$. Find a number $n$ which has   


*

*$12$ divisors $(1 = d_1 < d_2 < \cdots <d_{12}=n )$ and   

*$d_{d_4-1}=(d_1+d_2+d_4)d_8$.    


Note: This is a problem from Russian Mathematical Olympiad 1989.
I think this is related with this one. But this seems harder to me. 
 A: A shorter solution than my original posted here: 
As you can see, $d_1+d_2+d_4>d_4$. By other side, as $d_1+d_2+d_4$ divides a divisor of $n$, implies that $d_1+d_2+d_4=d_i$ for $4<i<12$. Note that 
$$d_1+d_2+d_4=\frac{d_{d_4-1}}{d_8} \leq \frac{n}{d_8}=d_5$$ 
So, as $d_4< d_1+d_2+d_4 \leq d_5$, implies the equality $d_1+d_2+d_4=d_5$.
So $d_{d_4-1}=n$, that means that $d_4-1=12$, so $d_4=13$.
Now, as the number has 12 divisors, it must be of the form $p^{11}$, $p^5q$, $p^3q^2$ $p^2qr$. 


*

*If $n=p^{11}$, $p=13$, but $d_4=13$, so $d_1=d_4$. Contradiction

*If $n=p^5q$ or $n=p^3q^2$, $p$ or $q$ must be 13 since $d_4=13$, so the other must be 2, or if not, $d_1+d_2+d_4$ will be even, and $n$ will have three prime divisors and that doesn't happen for this case. But, if the other prime is 2, the only valid sum able for $d_1+d_2+d_4$ can be $1+2+13=16$, and 16 divides $n$ only in the case where $n=p^5q$, with $p=2$, but if that happens, $d_4=8$, and makes a contradiction.
So the only valid case is $n=p^2qr$, with $d_2,d_3<13$. Again, 2 doesn't divide $n$ since $d_1+d_2+d_4$ will be equal to 16, and $p^4$ doesn't divide $n$ for any prime $p$. 
For $d_3\neq d_2^2$, we have that $d_2=3,5,7$, that implies that 
$$d_1+d_2+d_4= 1+3+13, 1+5+13, 1+7+13= 17,19,21$$
In the first two cases $n$ doesn't have possible values for $d_3$, and in the last one, 3 divides $n$, making $d_2=3=7=d_2$, contradiction.
So the only way is making $d_2=3,d_3=d_2^2=9$, and this lets me $d_1+d_2+d_4=1+3+13=17$, so $n=9*13*17=1989$, and is clear that $17*117=1989$
