Найдите остаток от деления на \(\displaystyle 71\) числа \(\displaystyle 1^{2025}+2^{2025}+3^{2025}+ \ldots +70^{2025}\small.\)
Имеем:
\(\displaystyle 70\equiv (-1)\hspace{-2mm}\pmod {71}\small.\)
По свойству сравнений
получаем
\(\displaystyle 70^{2025}\equiv (-1)^{2025}\hspace{-2mm}\pmod {71}\small.\)
Значит,
\(\displaystyle 1^{2025}+70^{2025}\equiv 1^{2025}+(-1)^{2025} \hspace{-2mm}\pmod {71}\small,\)
\(\displaystyle 1^{2025}+70^{2025} \equiv 0\hspace{-2mm}\pmod {71}\small.\)
Далее,
\(\displaystyle 69\equiv (-2)\hspace{-2mm}\pmod {71}\small,\)
\(\displaystyle 69^{2025}\equiv (-2)^{2025}\hspace{-2mm}\pmod {71}\small,\)
\(\displaystyle 2^{2025}+69^{2025}\equiv 2^{2025}+(-2)^{2025} \hspace{-2mm}\pmod {71}\small,\)
\(\displaystyle 2^{2025}+69^{2025} \equiv 0\hspace{-2mm}\pmod {71}\small.\)
Аналогично,
\(\displaystyle 3^{2025}+68^{2025} \equiv 0\hspace{-2mm}\pmod {71}\small,\)
\(\displaystyle 4^{2025}+67^{2025} \equiv 0\hspace{-2mm}\pmod {71}\small,\)
\(\displaystyle \ldots \)
\(\displaystyle 35^{2025}+36^{2025} \equiv 0\hspace{-2mm}\pmod {71}\small.\)
Перегруппируем слагаемые:
\(\displaystyle 1^{2025}+2^{2025}+3^{2025}+ \ldots +70^{2025}=\)
\(\displaystyle =\left(1^{2025}+70^{2025}\right)+\left(2^{2025}+69^{2025}\right)+ \ldots +\left(35^{2025}+36^{2025}\right)\small.\)
Тогда
\(\displaystyle \left(1^{2025}+70^{2025}\right)+\left(2^{2025}+69^{2025}\right)+ \ldots +\left(35^{2025}+36^{2025}\right)\equiv 0+0+\ldots +0\hspace{-2mm}\pmod {71}\small,\)
\(\displaystyle \left(1^{2025}+70^{2025}\right)+\left(2^{2025}+69^{2025}\right)+ \ldots +\left(35^{2025}+36^{2025}\right)\equiv 0\hspace{-2mm}\pmod {71}\small.\)
Следовательно, остаток от деления \(\displaystyle 1^{2025}+2^{2025}+3^{2025}+ \ldots +70^{2025}\hspace{-2mm}\small\) на \(\displaystyle 71\) равен \(\displaystyle 0\small.\)
Ответ: \(\displaystyle 0\small.\)