, 2015

Solution:

Answer: $153$

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 $1 + 1 + 1 + 0 + 150 = 153$.

Why is this optimal? Suppose $(a, b, c, d, e)$ maximizes $a + b + c + d + e$. Then $a, b, c, d \leq 1$, or else we could use a replacement $(a_{i}, a_{i+1}) \rightarrow (a_{i} - 2, a_{i+1} + 5)$ to strictly increase the sum. But modulo $2$ forces $a$ odd, so $a = 1$. Subtracting off and continuing in this manner shows that we must have $b = 1$, then $c = 1$, then $d = 0$, and finally $e = 150$.