Find the natural number(s) n with $12$ divisors $1=d_1<d_2<…<d_{12}=n$ such that the divisor with the index $d_4$, i.e, $d_{d_4}$ is equal to $1+(d_1+d_2+d_4)d_8$.
My work:
$$\begin{align}
&\implies d_{d_4} = 1+(d_1+d_2+d_4)d_8 \\
&\implies d_{d_4}-d_1 =(d_1+d_2+d_4)d_8 \\
&\implies d_1+d_2+d_4=\frac{d_{d_4}}{d_8}-\frac{d_1}{d_8} \lt\frac{n}{d_8}=d_5 \\
&\implies d_1+d_2+d_4<d_5
\end{align}$$
Again, as all divisors must be positive integers, we must have,
$$d_4<d_1+d_2+d_4<d_5$$
Also as $d_{d_4}$ is a divisor, $4\le d_4\le 12$, as 4 is the minimum possible value of $d_4$, which is attained when $d_2=2$ and $d_3=3$
I am stuck here. Any ideas on how to proceed or the solution will be greatly appreciated.
Best Answer
Where did you get this question? There are no solutions.
First, $d_8<d_{d_4}$, so $8<d_4$.
In general, $d_i d_{13-i}=n$. Substituting $i=d_4$ gives that $n/d_{d_4}=d_{13-d_4}$. Then $n/d_{13-d_4}=1+(d_1+d_2+d_4)d_8$, so multiplying by $d_{13-d_4}$ and taking it mod $d_8$ gives that $d_8|d_{13-d_4}$, so $8\leq 13-d_4$, $d_4\leq 5$.
This is a contradiction, so there are no such solutions.