Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 262

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\[1)\ 28 \cdot 20^{15} - 10 \cdot 18^{13}\ \ \vdots 19\]

\[28 \equiv 9(mod\ 19);\ \]

\[\ 20 \equiv 1(mod\ 19);\]

\[18 \equiv - 1(mod\ 19);\]

\[a = 9 \cdot 1^{15} - 10 \cdot ( - 1)^{13} =\]

\[= 9 + 10 = 19 \equiv 0(mod\ 19).\]

\[2)\ 3^{3} \cdot 23^{8} + 5^{8} \cdot 2^{14} =\]

\[= 27 \cdot 23^{8} + 10^{8} \cdot 2^{6}\ \vdots 13\]

\[3^{3} = 27 \equiv 1(mod\ 13);\]

\[23 \equiv 10(mod\ 13);2^{3} =\]

\[= 8 \equiv - 5(mod\ 13);\]

\[a = 1 \cdot 10^{8} + 10^{8} \cdot ( - 5)^{2} =\]

\[= 10^{8}(1 + 25) =\]

\[= 10^{8} \cdot 26 \equiv 0(mod\ 13).\]

\[3)\ 125^{24} - 1825 =\]

\[= \left( 25^{3} \right)^{24} - \underset{\vdots 600}{\overset{1800}{︸}} - 25\ \]

\[25^{2} = 625 \equiv 25(mod\ 600);\ \]

\[a = 25 - 25 \equiv 0(mod\ 600);\]

\[125^{24} - 1825\ \vdots 600.\]

\[4)\ 100^{20} - 50 \cdot 16^{5} =\]

\[= 2^{20} \cdot 50^{20} - 50 \cdot \left( 2^{4} \right)^{5} =\]

\[= 2^{20} \cdot 50^{20} - 50 \cdot 2^{20} =\]

\[= 2^{20} \cdot \left( 50^{20} - 50 \right)\]

\[50 \equiv 1(mod\ 49);\ \ \]

\[1^{20} - 1 = 1 - 1 = 0.\]

\[100^{20} - 50 \cdot 16^{5}\ \vdots 49.\]

\[5)\ 42^{365} + 53^{247}\ \ \vdots 5\]

\[42 \equiv 2(mod\ 5);53 \equiv 3(mod\ 5);\]

\[2 + 3(mod\ 5) = 0(mod\ 5).\]

\[6)\ 71^{325} + 41^{135}\ \vdots 7\]

\[71 \equiv 1(mod\ 7);\ \ \]

\[41 \equiv - 1(mod\ 7);\]

\[a = 1^{325} + ( - 1)^{135} =\]

\[= 1 - 1 \equiv 0(mod\ 7).\]