Question

CMR: 

2005ⁿ+60ⁿ-1897ⁿ-168ⁿ⋮2004

Answer 1

Đặt \(A=2005^n+60^n-1897^n-168^n\)

\(2004=4.3.167\)

2005 chia 4 dư 1 nên \(2005^n\equiv1\left(mod4\right)\)

\(1897\) chia 4 dư 1 nên \(1897^n\equiv1\left(mod4\right)\)

Tương tự: \(60^n\equiv0\left(mod4\right)\) ; \(168^n\equiv0\left(mod4\right)\)

\(\Rightarrow2005^n+60^n-1897^n-168^n\equiv1+0-1-0\equiv0\left(mod4\right)\)

\(\Rightarrow A⋮4\)

Cũng làm như vậy, ta có:

\(2005^n+60^n-1897^n-168^n\equiv1+0-1-0\equiv0\left(mod3\right)\)

\(\Rightarrow A⋮3\)

\(2005^n+60^n-1897^n-168^n\equiv1+60^n-60^n-1\equiv0\left(mod167\right)\)

\(\Rightarrow A⋮167\)

Mà 4, 3, 167 nguyên tố cùng nhau

\(\Rightarrow A⋮\left(4.3.167\right)\) hay \(A⋮2004\)