Найдите остаток от деления на \(\displaystyle 6\) числа \(\displaystyle 201^{2025}+204^{2025}\small.\)
Имеем:
\(\displaystyle 201\equiv 3\hspace{-2mm}\pmod {6}\small.\)
\(\displaystyle 204\equiv 0\hspace{-2mm}\pmod {6}\small.\)
По свойству сравнений
получаем
\(\displaystyle 201^{2025}\equiv 3^{2025}\hspace{-2mm}\pmod {6}\small.\)
\(\displaystyle 204^{2025}\equiv 0^{2025}\hspace{-2mm}\pmod {6}\small.\)
Заметим, что
\(\displaystyle 3^{2}\equiv 3\hspace{-2mm}\pmod {6}\small,\)
\(\displaystyle 3^{3}=3^{2}\cdot 3\small,\) \(\displaystyle 3^{3}\equiv 3\cdot 3\hspace{-2mm}\pmod {6}\small,\) \(\displaystyle 3^{3}\equiv 3\hspace{-2mm}\pmod {6}\small,\)
\(\displaystyle 3^{4}=3^{3}\cdot 3\small,\) \(\displaystyle 3^{4}\equiv 3\cdot 3\hspace{-2mm}\pmod {6}\small,\) \(\displaystyle 3^{4}\equiv 3\hspace{-2mm}\pmod {6}\small,\)
\(\displaystyle \ldots\)
\(\displaystyle 3^{2025}=3^{2024}\cdot 3\small,\) \(\displaystyle 3^{2025}\equiv 3\cdot 3\hspace{-2mm}\pmod {6}\small,\) \(\displaystyle 3^{2025}\equiv 3\hspace{-2mm}\pmod {6}\small.\)
Тогда
\(\displaystyle 201^{2025}+204^{2025}\equiv 3^{2025}+0^{2025} \hspace{-2mm}\pmod {6}\small,\)
\(\displaystyle 201^{2025}+204^{2025}\equiv 3+0 \hspace{-2mm}\pmod {6}\small,\)
\(\displaystyle 201^{2025}+204^{2025} \equiv 3\hspace{-2mm}\pmod {6}\small.\)
Значит, остаток от деления \(\displaystyle 201^{2025}+204^{2025} \small\) на \(\displaystyle 6\) равен \(\displaystyle 3\small.\)
Ответ: \(\displaystyle 3\small.\)