回復 1# whzzthr 的帖子
另解
\(\begin{align}
& {{2}^{3}}\equiv -1\ \left( \bmod \ 9 \right) \\
& {{2}^{101}}={{2}^{2}}\times {{\left( {{2}^{3}} \right)}^{33}}\equiv 4\times {{\left( -1 \right)}^{33}}\equiv -4\equiv 5\ \left( \bmod \ 9 \right) \\
& \\
& {{3}^{101}}\equiv 0\ \left( \bmod \ 9 \right) \\
& \\
& {{5}^{3}}\equiv -1\ \left( \bmod \ 9 \right) \\
& {{5}^{101}}={{5}^{2}}\times {{\left( {{5}^{3}} \right)}^{33}}\equiv 7\times {{\left( -1 \right)}^{33}}\equiv -7\equiv 2\ \left( \bmod \ 9 \right) \\
& \\
& {{2}^{101}}+{{3}^{101}}+{{5}^{101}}\equiv 5+0+2\equiv 7\ \left( \bmod \ 9 \right) \\
\end{align}\)
