We have given a natural number $n$ with divisors $(1) \;\; 1 = d_1 < d_2 < \cdots < d_k = n$ satisfying the equation $(2) \;\; n = d_2d_3 + d_2d_5 + d_3d_5$. Now $d_2=p$, where $p$ is the smallest prime divisor of $n$. If $n$ has more than one prime divisor, let $q$ be the second smallest of these prime divisors. Then we have two cases to consider: Case 1: $(d_2,d_3) = (p,p^2)$. Then $d_5 \mid d_2d_3 = p^3$ by equation (2), yielding $d_5=p^3$ since $d_5 > p^2 = d_3$. Hence by equation (2) $(3) \;\; n = p^3(p^2 + p + 1)$, yielding $n < p^6 = d_5^2$, which give us ${\textstyle d_5 > \sqrt{n} > d_{[k/2]}}$, yielding $k < 10$. If $p^2 + p + 1$ is composite, then $k \geq 4 \cdot 3 = 12$ by equation (3). Hence $p^2 + p + 1 = q$ and $k = (3 + 1)(1 + 1) = 8$ by equation (3). Case 2: $(d_2,d_3) = (p,q)$. Then $d_5 \mid d_2d_3 = pq$, yielding $d_5=pq$ since $d_5 > d_3 = q > d_2 = p$. Hence by equation (2) $(4) \;\; n = pq(1 + p + q)$. Therefore, since $d_3 < d_4 < d_5$, we obtain $(5) \;\; q < d_4 < pq$ The fact that $q < 1 + p + q < pq$ is equivalent $(p - 1)(q - 1) > 2$ combined with equation (4) and inequality (5) implies $d_4 = p + q + 1$ when $(p,q) \neq (2,3)$. Consequently $n = d_5d_4$ by equation (4), which implies $k=4+5-1=8=2^3$. Hence $p + q + 1$ must be a prime by equation (4). Setting $(p,q)=(2,3)$ in equation (4), we obtain the solution $n = 2^2 \cdot 3^2$ and $k = (2 + 1)^2 = 9$. Conclusion: The natural numbers $n$ satisfying condition (1) and equation (2) have 8 or 9 divisors, i.e. $k \in \{8,9\}$.

Solar Plexsus wrote: We have given a natural number $n$ with divisors $(1) \;\; 1 = d_1 < d_2 < \cdots < d_k = n$ satisfying the equation $(2) \;\; n = d_2d_3 + d_2d_5 + d_3d_5$. Now $d_2=p$, where $p$ is the smallest prime divisor of $n$. If $n$ has more than one prime divisor, let $q$ be the second smallest of these prime divisors. Then we have two cases to consider: Case 1: $(d_2,d_3) = (p,p^2)$. Then $d_5 \mid d_2d_3 = p^3$ by equation (2), yielding $d_5=p^3$ since $d_5 > p^2 = d_3$. Hence by equation (2) $(3) \;\; n = p^3(p^2 + p + 1)$, yielding $n < p^6 = d_5^2$, which give us ${\textstyle d_5 > \sqrt{n} > d_{[k/2]}}$, yielding $k < 10$. If $p^2 + p + 1$ is composite, then $k \geq 4 \cdot 3 = 12$ by equation (3). Hence $p^2 + p + 1 = q$ and $k = (3 + 1)(1 + 1) = 8$ by equation (3). Case 2: $(d_2,d_3) = (p,q)$. Then $d_5 \mid d_2d_3 = pq$, yielding $d_5=pq$ since $d_5 > d_3 = q > d_2 = p$. Hence by equation (2) $(4) \;\; n = pq(1 + p + q)$. Therefore, since $d_3 < d_4 < d_5$, we obtain $(5) \;\; q < d_4 < pq$ The fact that $q < 1 + p + q < pq$ is equivalent $(p - 1)(q - 1) > 2$ combined with equation (4) and inequality (5) implies $d_4 = p + q + 1$ when $(p,q) \neq (2,3)$. Consequently $n = d_5d_4$ by equation (4), which implies $k=4+5-1=8=2^3$. Hence $p + q + 1$ must be a prime by equation (4). Setting $(p,q)=(2,3)$ in equation (4), we obtain the solution $n = 2^2 \cdot 3^2$ and $k = (2 + 1)^2 = 9$. Conclusion: The natural numbers $n$ satisfying condition (1) and equation (2) have 8 or 9 divisors, i.e. $k \in \{8,9\}$. I have good conclusions We know that $d_i*d_(k-i+1)=n$ Then we get $p^3*(p^2+p+1)=n$ So d5*d(k-4)=n Which is d5>dk-4 Then 5>k-4 9>k>5 Try k=(6,7,8) we can easliy obtain that And the second case is similiar with it