Finding natural numbers with $12$ divisors $1=d_1<d_2<\cdots<d_{12}=n$, such that the divisor with the index $d_4$ is equal to $1+(d_1+d_2+d_4)d_8$.

Question

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.

Answer 1

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.

Answer 2

I will be slightly cheeky here and give you a partial answer... but one that should be easy to finish off.

Write $n=p_1^{k_1}p_2^{k_2}\dots p_l^{k_l}$ with $p_1 < \dots < p_l$ primes. Then the number of divisors of $n$ is equal to $(k_1+1)\dots (k_l+1)$. This implies that $l$ can be at most $3$ and $(k_1,k_2,k_3)$ must be one of the following:

$$\{(11,0,0),(1,5,0),(5,1,0),(2,3,0),(3,2,0),(1,1,2),(1,2,1),(2,1,1)\}.$$

Now, unfortunately, I don't see anything better to do than consider each of the $8$ cases in turn... I will do a couple.

First, the easiest of the cases. That is, $(k_1,k_2,k_3)=(11,0,0)$ i.e. $n=p^{11}$ for some prime $p$. Then listing the divisors in order gives us \begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|} \hline d_1& d_2 & d_3 & d_4 & d_5 & d_6 &d_7& d_8 & d_9 & d_{10} & d_{11} & d_{12} \\ \hline 1&p&p^2&p^3&p^4&p^5&p^6&p^7&p^8&p^9&p^{10}&p^{11}\\ \hline \end{array} Now we require $d_{d_4} = 1 + (d_1+d_2+d_4)d_8 \iff d_{p^3} = 1 + (1+p+p^3)p^7$. However, since $d_{p^3}$ must be a divisor we must have $p^3 \in \{4,5,...,12\}$. This implies $p=2$, and $d_{p^3} = d_8 = p^7$. This clearly doesn't satisfy the condition so we rule out the $(k_1,k_2,k_3) = (11,0,0)$ case.

Another case: $(k_1,k_2,k_3)=(1,1,2)$ i.e. $n=p_1p_2p_3^{2}$. Then the list of divisors is: \begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|} \hline d_1& d_2 & d_3 & - & - & d_6 &d_7& - & - & d_{10} & d_{11} & d_{12} \\ \hline 1&p_1&p_2&p_3&p_1p_2&p_1p_3&p_2p_3&p_3^2&p_1p_2p_3&p_1p_3^2&p_2p_3^3&p_1p_2p_3^2\\ \hline \end{array}

Note that I haven't labelled $d_4$ or $d_5$ (and hence $d_8$ or $d_9$) because we don't know whether $p_3$ or $p_1p_2$ is bigger. So we consider both cases seperately.

Case a): $p_3 < p_1p_2$ i.e. $d_4 = p_3, d_5 = p_1p_2$ (and hence $d_8 = p_3^2, d_9 = p_1p_2p_3).$

In this case we require $d_{d_4} = d_{p_3} = 1 + (1+p_1+p_3)p_3^2$. However, note that the RHS is greater than $p_3^3$ which itself is greater than $n$. So the divisor condition can never be met.

Case b): $p_3 > p_1p_2$ i.e. $d_4 = p_1p_2, d_5 = p_3$.

In this case we must have $p_1p_2 \in \{4,5,...,12\}$ (since $d_{p_1p_2}$ must be a divisor) and so $p_1p_2$ is either $6$ or $10$.

If $p_1p_2 = 6$, then we require $d_6 = p_1p_3 = 1 + (1+p_1+p_1p_2)p_1p_2p_3$ - impossible.

Finally if $p_1p_2 = 10$ then we require $d_{10} = p_1p_3^2 = 1 + (1+p_1+p_1p_2)p_1p_2p_3$. You can rule this out too by considering the condition modulo $p_1$.

Therefore $(k_1,k_2,k_3)=(1,1,2)$ yields no solutions too.

I will admit that I have not done all the other cases, but I am fairly confident you will be able to get a full answer by continuing in this way. Although hopefully someone can give you a more elegant method.