The intuition is that as much should be in $e$ as possible. But divisibility obstructions like $16 \nmid 15^{3}$ are in our way. However, the way the coefficients $5^{4}>5^{3} \cdot 2>\cdots$ are set up, we can at least easily avoid having $a, b, c, d$ too large (specifically, $\geq 2$ ). This is formalized below. First, we observe that $(a_{1}, a_{2}, a_{3}, a_{4}, a_{5})=(5,1,0,0,0)$ is a solution. Then given a solution, replacing $(a_{i}, a_{i+1})$ with $(a_{i}-2, a_{i+1}+5)$, where $1 \leq i \leq 4$, also yields a solution. Given a solution, it turns out all solutions can be achieved by some combination of these swaps (or inverses of these swaps). Thus, to optimize the sum, we want $(a, b, c, d) \in\{0,1\}^{4}$, since in this situation, there would be no way to make swaps to increase the sum. So the sequence of swaps looks like $(5,1,0,0,0) \rightarrow(1,11,0,0,0) \rightarrow (1,1,25,0,0) \rightarrow(1,1,1,60,0) \rightarrow(1,1,1,0,150)$, yielding a sum of $1+1+1+0+150=153$.