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Question Number 203192 by MrGHK last updated on 12/Jan/24

$$\mathbf{find}\mathbf{the}\mathbf{last}\mathbf{four}\mathbf{digits}\mathbf{of}{2022}^{2023}+{2023}^{2022}$$

Answered by deleteduser1 last updated on 12/Jan/24

$$x={2022}^{2023}+{2023}^{2022}\stackrel{16}{\equiv }{6}^{2023}+7^{2022}\stackrel{16}{\equiv }{2}^{2023}{3}^{2023}+7^6$$$$\stackrel{16}{\equiv }{7}^6\stackrel{16}{\equiv }1$$$$x={2022}^{2023}+{2023}^{2022}\stackrel{625}{\equiv }{\left(147\right)}^{23}+{148}^{22}\stackrel{625}{\equiv }$$$$\stackrel{625}{\equiv }{\left(147\right)}^{26}{\left(147\right)}^{-2}+{\left(148\right)}^{24}{\left(148\right)}^{-2}$$$$\equiv 4{\left(147\right)}^{-2}+356{\left(148\right)}^{-2}$$$$\frac{1}{{148}^2}\stackrel{625}{\equiv }z\Rightarrow {148}^{2}z\equiv 29z\stackrel{625}{\equiv }1\Rightarrow z\stackrel{625}{\equiv }194$$$$\Rightarrow x\equiv 4\left(372\right)+356\left(194\right)\equiv 552\left(mod625\right)$$$$x=625q+552\equiv 1\left(mod16\right)\Rightarrow q\equiv 9\left(mod16\right)$$$$\Rightarrow x=625(16k+9)+552\Rightarrow x=10000k+6177$$

Commented by MrGHK last updated on 13/Jan/24

$$thanks$$

Commented by MathematicalUser2357 last updated on 15/Jan/24

$$q+552$$$$readtheblueone$$

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