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find-the-last-4-digits-of-2024-2023-




Question Number 203270 by MrGHK last updated on 13/Jan/24
find the last 4 digits of 2024^(2023)
$$\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{last}}\:\mathrm{4}\:\boldsymbol{\mathrm{digits}}\:\boldsymbol{\mathrm{of}}\:\mathrm{2024}^{\mathrm{2023}} \\ $$
Answered by Frix last updated on 14/Jan/24
Last 4 digits of 2024^n   n=1 2024  Then a loop of length 50  f(n)=last 4 digits of 2024^n   f(n)=f(n+50k); 2≤n≤51∧k∈N  f(2023)=f(23)=7824
$$\mathrm{Last}\:\mathrm{4}\:\mathrm{digits}\:\mathrm{of}\:\mathrm{2024}^{{n}} \\ $$$${n}=\mathrm{1}\:\mathrm{2024} \\ $$$$\mathrm{Then}\:\mathrm{a}\:\mathrm{loop}\:\mathrm{of}\:\mathrm{length}\:\mathrm{50} \\ $$$${f}\left({n}\right)=\mathrm{last}\:\mathrm{4}\:\mathrm{digits}\:\mathrm{of}\:\mathrm{2024}^{{n}} \\ $$$${f}\left({n}\right)={f}\left({n}+\mathrm{50}{k}\right);\:\mathrm{2}\leqslant{n}\leqslant\mathrm{51}\wedge{k}\in\mathbb{N} \\ $$$${f}\left(\mathrm{2023}\right)={f}\left(\mathrm{23}\right)=\mathrm{7824} \\ $$

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