An equation involving divisors

Let $p_m$ denote the $m$ th prime number. Then we can write the prime factorization of $n$ as $n = \prod_{m=1}^M p_m^{a_m}$ for some positive integer $M$ and non-negative integers $a_m$ for $m=1,2,\ldots, M.$ Then the given equation becomes $\left(\prod_{m=1}^M (a_m+1)\right)^3 = 4n = 2^{a_1+2}\cdot 3^{a_2}\cdot 5^{a_3}\cdot \prod_{m=4}^M p_m^{a_m}$ Since the left-hand side is a perfect cube, the right-hand side must be as well, so each of the exponents in the prime factorization must be a multiple of $3.$ Therefore, there are non-negative integers $b_m$ such that $a_1 = 3b_1+1$ and $a_m = 3b_m$ for all $m=2,3,\ldots, M,$ giving $\left((3b_1+2)(3b_2+1)(3b_3+1)\prod_{m=4}^M (3b_m+1)\right)^3 = 2^{3b_1+3} \cdot 3^{3b_2} \cdot 5^{3b_3} \cdot \prod_{m=4}^M p_m^{3b_m}$ $(3b_1+2)(3b_2+1)(3b_3+1)\prod_{m=4}^M (3b_m+1) = 2^{b_1+1} \cdot 3^{b_2} \cdot 5^{b_3} \cdot \prod_{m=4}^M p_m^{b_m}$ From here, we note that the left-hand side isn't divisible by $3$ , so the right-hand side isn't divisible by $3$ either, so $b_2=0,$ now yielding $(3b_1+2)(3b_3+1)\prod_{m=4}^M (3b_m+1) = 2^{b_1+1} \cdot 5^{b_3} \cdot \prod_{m=4}^M p_m^{b_m}$ $\frac{2^{b_1+1}}{3b_1+2} \cdot \frac{5^{b_3}}{3b_3+1} = \left(\prod_{m=4}^M \frac{p_m^{b_m}}{3b_m+1}\right)^{-1}$

We can check that the minimum of the left-hand side of this equation occurs when $b_1=1, b_3=0$ so $\frac{4}{5} \leq \left(\prod_{m=4}^M \frac{p_m^{b_m}}{3b_m+1}\right)^{-1} \\ \implies \prod_{m=4}^M \frac{p_m^{b_m}}{3b_m+1} \leq \frac{5}{4}$ Obviously this is satisfied when all the $b_m = 0$ for $m=4,5,\ldots,M$ , but if we assume that at least one of them is nonzero, then the minimum value would occur when $b_4=1$ and $b_m=0$ for $m=5,6,\ldots,M$ , giving $\frac{7}{4} = \frac{7^1}{3(1)+1} \prod_{m=5}^M \frac{p_m^0}{3(0)+1} \leq \prod_{m=4}^M \frac{p_m^{b_m}}{3b_m+1} \leq \frac{5}{4}$ which is a contradiction, so we must have $b_m=0$ for all $m\geq 4$ .

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Going back to a previous equation, we now want to solve $(3b_1+2)(3b_3+1) = 2^{b_1+1} \cdot 5^{b_3}$ which we can do just by cases with $b_3$ as follows:

• Case $b_3=0$ : Then $3b_1+2 = 2^{b_1+1}$ . We can check that $b_1=0,2$ are solutions and $b_1=1$ is not a solution just by plugging in those values. Further, for $b_1 > 2$ we have $2^{b_1+1} > 3b_1+2$ , so the only two solutions are $b_1=0,2.$
• Case $b_3=1$ : Then $3b_1+2 = 5 \cdot 2^{b_1-1}$ , then similarly, we can see that $b_1=0$ is not a solution and $b_1=1$ is a solution by plugging values in. Further, for $b_1>1$ we have $5\cdot 2^{b_1-1} > 3b_1+2$ , so the only solution is $b_1=1.$
• Case $b_3>1$ : Then $\left(\frac{2^{b_1+1}}{3b_1+2}\right)^{-1} = \frac{5^{b_3}}{3b_3+1} > \frac{5}{4} \\ \implies \frac{2^{b_1+1}}{3b_1+2} < \frac{4}{5}$ but as our previous analysis implied, $\frac{2^{b_1+1}}{3b_1+2} \geq \frac{4}{5},$ so this is a contradiction, and there is no solution in this case.

Therefore we have shown three solutions, given by $(b_1,b_3) \in \{(0,0), (2,0), (1,1)\}$ and all other $b_m =0$ . Since then $n = 2^{3b_1+1} \cdot 5^{3b_3}$ which means the answer is $2^{3(0)+1}\cdot 5^{3(0)} + 2^{3(2)+1}\cdot 5^{3(0)} + 2^{3(1)+1}\cdot 5^{3(1)} = 2 + 128 + 2000 = \boxed{2130}$