Chứng minh rằng:a) \({4^{2018}} - 7 \vdots 9\)b) \({555^{222}} + {222^{555}}\) chia hết cho
Chứng minh rằng:
a) \({4^{2018}} - 7 \vdots 9\)
b) \({555^{222}} + {222^{555}}\) chia hết cho 7
Khi số dư trong phép chia \(a\) cho \(m\) bằng 0 thì \(a \vdots m\).
Như vậy để chứng tỏ \(a \vdots m\) ta chứng minh \(a \equiv 0(\,\bmod \,m)\)
a) Ta có \(\left\{ \begin{array}{l}{4^3} = 64 \equiv 1(\,\bmod \,9) \Rightarrow {4^{2016}} = {\left( {{4^3}} \right)^{672}} \equiv 1(\,\bmod \,9)\\{4^2} = 16 \equiv 7(\,\bmod \,9)\end{array} \right.\)
\( \Rightarrow {4^{2018}} = {4^{2016}}{.4^2} \equiv 1.7(\,\bmod \,9)\)
Vậy \({4^{2018}} - 7 \equiv 0(\,\bmod \,9)\) hay \({4^{2018}} - 7 \vdots 9\)
b) Ta có \(\left\{ \begin{array}{l}555 \equiv 2(\,\bmod \,7) \Rightarrow {555^{222}} \equiv {2^{222}}(\,\bmod \,7)\\{2^3} \equiv 1(\,\bmod \,7) \Rightarrow {2^{222}} = {\left( {{2^3}} \right)^{74}} \equiv 1(\,\bmod \,7)\end{array} \right.\)
\( \Rightarrow {555^{222}} \equiv 1(\,\bmod \,7)\)
Lại có: \(\left\{ \begin{array}{l}222 \equiv - 2(\,\bmod \,7) \Rightarrow {222^{555}} \equiv {( - 2)^{555}}(\,\bmod \,7)\\{( - 2)^3} \equiv - 1(\,\bmod \,7) \Rightarrow {( - 2)^{555}} = {\left[ {{{( - 2)}^3}} \right]^{185}} \equiv - 1(\,\bmod \,7)\end{array} \right.\)
\( \Rightarrow {222^{555}} \equiv - 1(\,\bmod \,7)\)
\( \Rightarrow {555^{222}} + {222^{555}} \equiv 0(\,\bmod \,7)\)
Vậy \({555^{222}} + {222^{555}}\) chia hết cho 7
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