= Solution
The <neutron> has valence content $udd$ and total <spin> $J=1/2$; its ground-state parity is positive. The two <down quarks> contribute charge $-2/3$ and the <up quark> contributes $+2/3$, giving net charge zero. “Three valence <quarks>” specifies the net flavor quantum numbers, rather than excluding <gluons> or additional sea <quark>–<antiquark> components from the full <Quantum chromodynamics> state.
In the approximate light-flavor SU(3) classification, the <neutron> belongs to the ground-state <baryon octet>, containing eight <hadrons>:
$$
\boxed{p,n;\quad\Sigma^+,\Sigma^0,\Sigma^-;\quad\Lambda^0;\quad\Xi^0,\Xi^-.}
$$
These are eight particle species, without counting their <antiparticles> again as extra members of the same octet. The corresponding <antibaryons> form the conjugate particle set. Within the smaller SU(2) <isospin> classification, the <neutron> instead belongs to the two-member nucleon doublet $(p,n)$. Stating the symmetry distinguishes the usual eight-member flavor multiplet answer from the also valid two-member <isospin> multiplet. The <baryon decuplet> has different ground-state spin and is not the <neutron>'s octet.
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