For a finite abelian group $G$ and $k \geq 1$, the generalized Davenport constant $D_k(G)$ is the least $\ell$ such that every sequence over $G$ of length at least $\ell$ has $k$ pairwise disjoint nonempty zero-sum subsequences. A theorem of Freeze and Schmid gives $D_k(C_5^3) \geq 5k+10$ for every $k \geq 2$. We prove the matching upper bound: $D_4(C_5^3)=30$, and hence $D_k(C_5^3)=5k+10$ for every $k \geq 2$, so the Freeze--Schmid bound is attained by $C_5^3$ from $k=2$ onward, as it is by $C_2^3$ and unlike $C_3^3$. The proof is finite and computer-assisted. The remaining case reduces to showing that every zero-sum sequence of length $31$ over $C_5^3$ contains a nonempty zero-sum subsequence of length at most five. A saturation argument confines the multiplicities of a hypothetical counterexample to $\{1,2,4\}$, its support pattern to one of $60$ solutions of two linear equations, and its geometry to one of $78$ rank/plane branches normalized to a standard basis; an exhaustive search exhausts every branch with no survivor. The search was carried out by three independently written implementations, and the branch cover was regenerated by separate programs from the lemmas alone; two further machine-verified values, $D_3(C_5^3)=25$ and $s_{\leq 6}(C_5^3)=24$, enter the second statement, and their records accompany the paper.

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