Question

A = \(27309^{10}+27309^{20}+27309^{30}+...+27309^{100}\)

Tìm số dư của A khi chia cho 7

Answer 1

Lời giải:

Ta thấy: \(27309\equiv 2\pmod 7\)

\(\Rightarrow A\equiv 2^{10}+2^{20}+2^{30}+...+2^{100}\pmod 7\)

Lại có:

\(2^3\equiv 1\pmod 7\)

\(\Rightarrow 2^{10}=(2^3)^3.2\equiv 1^3.2\equiv 2\pmod 7\)

\(\Rightarrow \left\{\begin{matrix} 2^{20}\equiv 2^2\pmod 7\\ 2^{30}\equiv 2^3\pmod 7\\ ......\\ 2^{100}\equiv 2^{10}\pmod 7\end{matrix}\right.\)

Do đó: \(A\equiv 2+2^2+..+2^{10}\pmod 7\)

\(A\equiv 2(1+2+2^2)+2^4(1+2+2^2)+2^7(1+2+2^2)+2^{10}\pmod 7\)

\(A\equiv 2.7+2^4.7+2^7.7+2^{10}\pmod 7\)

\(A\equiv 2^{10}\equiv 2\pmod 7\)

Vậy $A$ chia $7$ dư $2$