${3^{1000}}$ 在十进制中最后 $4$ 位是 \((\qquad)\) ．
【难度】
【出处】
2005年复旦大学保送生招生测试
【标注】
【答案】
A
【解析】
因为\[\begin{split}{3^{1000}} &= {9^{500}}= {\left( {10 - 1} \right)^{500}}\\& = {\mathrm {C}}_{500}^0{10^{500}} + {\mathrm {C}}_{500}^{1}{10^{499}}{\left( { - 1} \right)^{1}}+ \cdots+ {\mathrm {C}}_{500}^{498}{10^2}{\left( { - 1} \right)^{498}} + {\mathrm {C}}_{500}^{499}{10^1}{\left( { - 1} \right)^{499}} + {\mathrm {C}}_{500}^{500}{10^0}{\left( { - 1} \right)^{500}},\end{split}\]而后 $4$ 位数字由$${\mathrm {C}}_{500}^{498}{10^2}{\left( { - 1} \right)^{498}} + {\mathrm {C}}_{500}^{499}{10^1}{\left( { - 1} \right)^{499}} + {\mathrm {C}}_{500}^{500}{10^0}{\left( { - 1} \right)^{500}}$$决定，而$${\mathrm {C}}_{500}^{498}{10^2}{\left( { - 1} \right)^{498}} + {\mathrm {C}}_{500}^{499}{10^1}{\left( { - 1} \right)^{499}} + {\mathrm {C}}_{500}^{500}{10^0}{\left( { - 1} \right)^{500}} = 12370001,$$所以最后 $4$ 位是 $0001$．
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