Question Number 227154 by gregori last updated on 03/Jan/26
$$ \left(\mathrm{202519}\right)^{\mathrm{2025}} \: \\ $$$$\: \\ $$
Answered by mahdipoor last updated on 03/Jan/26
$$\mathrm{202519}\equiv−\mathrm{2}\:\:\:\mathrm{mod}\left(\mathrm{17}\right) \\ $$$$\left(−\mathrm{2}\right)^{\mathrm{5}} \equiv\mathrm{2}\:\Rightarrow\:\left(\mathrm{2}\right)^{\mathrm{5}} \equiv−\mathrm{2}\:\:\Rightarrow\:\left(−\mathrm{2}\right)^{\mathrm{25}} \equiv−\mathrm{2} \\ $$$$\left(−\mathrm{2}\right)^{\mathrm{9}} \equiv−\mathrm{2}\:\Rightarrow\:\left(−\mathrm{2}\right)^{\mathrm{81}} \equiv−\mathrm{2} \\ $$$$\Rightarrow\left(−\mathrm{2}\right)^{\mathrm{2025}} =\left(−\mathrm{2}\right)^{\mathrm{25}×\mathrm{81}} \equiv\left(−\mathrm{2}\right)^{\mathrm{81}} \equiv−\mathrm{2}\equiv\mathrm{15} \\ $$
Answered by AgniMath last updated on 04/Jan/26
$${gcd}\left(\mathrm{202519},\mathrm{17}\right)=\mathrm{1} \\ $$$${according}\:{to}\:{carmichael}\:{theorem}\:: \\ $$$$\lambda\left(\mathrm{17}\right)=\mathrm{16} \\ $$$${Reduce}\:{base}\:{and}\:{exponent} \\ $$$$\left(\mathrm{202519}\right)^{\mathrm{2025}} \:{mod}\:\mathrm{17}\:\equiv\:\left(−\mathrm{2}\right)^{\mathrm{9}} \:{mod}\:\mathrm{17} \\ $$$$\equiv\:−\mathrm{2}+\mathrm{17}=\mathrm{15} \\ $$
Answered by mehdee7396 last updated on 04/Jan/26
$$\mathrm{202519}\underset{\mathrm{17}} {\equiv}−\mathrm{2}\Rightarrow\left(\mathrm{202519}\right)^{\mathrm{4}} \underset{\mathrm{27}} {\equiv}−\mathrm{1} \\ $$$$\left(\mathrm{202519}\right)^{\mathrm{2024}} \underset{\mathrm{17}} {\equiv}\mathrm{1}\Rightarrow\left(\mathrm{201519}\right)^{\mathrm{2025}} \underset{\mathrm{17}} {\equiv}−\mathrm{2}\underset{\mathrm{17}} {\equiv}\mathrm{15} \\ $$