Question Number 184076 by liuxinnan last updated on 02/Jan/23
$$\mathrm{6647}^{\mathrm{3}} {mod}\mathrm{10000}=\mathrm{2023} \\ $$
Commented by liuxinnan last updated on 02/Jan/23
$${maybe}\:{there}\:{are}\:{other}\:{number} \\ $$$${a}^{{n}} {mod}\mathrm{10000}=\mathrm{2023} \\ $$$${n}\in\mathbb{N}\:{n}\geqslant\mathrm{3} \\ $$$${find}\:{more}\:{a}\:{and}\:{n} \\ $$
Commented by Rasheed.Sindhi last updated on 03/Jan/23
$$\left.\mathrm{Surely}\:\mathrm{2023}^{\mathrm{1}} \mathrm{mod}\:\mathrm{10000}=\mathrm{2023}\::\right) \\ $$$$\mathrm{Happy}\:\mathrm{New}\:\mathrm{Year}\:\mathrm{sir}! \\ $$