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Question Number 203270 by MrGHK last updated on 13/Jan/24

$$\mathbf{find}\mathbf{the}\mathbf{last}4\mathbf{digits}\mathbf{of}{2024}^{2023}$$

Answered by Frix last updated on 14/Jan/24

$$\mathrm{Last}4\mathrm{digits}\mathrm{of}{2024}^n$$$$n=12024$$$$\mathrm{Then}\mathrm{a}\mathrm{loop}\mathrm{of}\mathrm{length}50$$$$f\left(n\right)=\mathrm{last}4\mathrm{digits}\mathrm{of}{2024}^n$$$$f\left(n\right)=f(n+50k);2⩽n⩽51\wedge k\in \mathbb{N}$$$$f\left(2023\right)=f\left(23\right)=7824$$

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