= Solution
Use <strong induction> with three consecutive base cases:
$$
8=3+5,\qquad9=3\cdot3,\qquad10=5\cdot2.
$$
For $n\ge11$, assume all values from $8$ through $n-1$ have the required representation. Since $8\le n-3<n$, write $n-3=3k+5m$ with $k,m$ <nonnegative integers>. Then $n=3(k+1)+5m$ is another such representation. <Strong induction> proves the claim for every $n\ge8$. This is a <consecutive-base induction for additive semigroups>: the three base cases handle all residues before repeated addition of $3$.
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