求\(\displaystyle 2^{2023}\)除以1000的餘數
先考慮\(\displaystyle 2^{2020}\)除以125的餘數
可知\(\displaystyle 2^{2020} \equiv (1024)^2 \equiv 576 (mod\ 125) \)
因此\(\displaystyle 2^{2023} \equiv 608 (mod\ 1000) \)

所求為\(\displaystyle 2^{2023}-2 \equiv 608-2 \equiv 606 (mod\ 1000) \)

黎曼和參考樓上PDEMAN老師

\(\displaystyle u=\sqrt{1+4x^2}  \Rightarrow u^2=1+4x^2  \Rightarrow \frac{1}{4}udu=xdx\)
原積分式為\(\displaystyle \int_1^{\sqrt{5}} \frac{3}{4}du =\frac{3}{4}(\sqrt{5}-1)\)