Question

[Lê Hoàng Minh]

10^2020+10^2019+10^2018+10^2017+8 chia hết 24

Answer 1

[Yuri Sweet[𝕿𝖊𝖆𝖒 𝕹𝖊𝖕𝖆𝖑]]

\(10^{2020}+10^{2019}+10^{2018}+10^{2017}+8⋮24\)

Giải

Ta có : \(10\equiv2\left(mod8\right)\)

\(\Rightarrow10^{2020}\equiv2^{2020}\left(mod8\right)\)

Ta lại có : \(2^3=8\equiv0\left(mod8\right)\)

\(\Rightarrow\left(2^3\right)^{673}\equiv0^{673}=0\left(mod8\right)\)

\(\Rightarrow\left(2^3\right)^{673}.2\equiv0.2=0\left(mod8\right)\)

hay \(10^{2020}\equiv0\left(mod8\right)\)

Tương tự ,ta được : \(10^{2019}\equiv0\left(mod8\right)\)

\(10^{2018}\equiv0\left(mod8\right)\)

\(10^{2017}\equiv0\left(mod8\right)\)

\(\Rightarrow10^{2020}+10^{2019}+10^{2018}+10^{2017}\equiv0\left(mod8\right)\)

mà \(8\equiv0\left(mod8\right)\)

\(\Rightarrow10^{2020}+10^{2019}+10^{2018}+10^{2017}+8\equiv0\left(mod8\right)\)

\(\Rightarrow10^{2020}+10^{2019}+10^{2018}+10^{2017}+8⋮8\left(1\right)\)

Mặt khác : \(10\equiv1\left(mod3\right)\)

\(\Rightarrow10^{2020}\equiv1^{2020}=1\left(mod3\right)\)

\(10^{2019}\equiv1^{2019}=1\left(mod3\right)\)

\(10^{2018}\equiv1^{2018}=1\left(mod3\right)\)

\(10^{2017}\equiv1^{2017}=1\left(mod3\right)\)

\(\Rightarrow10^{2020}+10^{2019}+10^{2018}+10^{2017}\equiv1+1+1+1=4\equiv1\left(mod3\right)\)

mà \(8\equiv2\left(mod3\right)\)

\(\Rightarrow10^{2020}+10^{2019}+10^{2018}+10^{2017}+8\equiv1+2=3\equiv0\left(mod3\right)\)

\(\Rightarrow10^{2020}+10^{2019}+10^{2018}+10^{2017}+8⋮3\left(2\right)\)

Từ \(\left(1\right),\left(2\right)\)và \(\left(8;3\right)=1\):

\(\Rightarrow10^{2020}+10^{2019}+10^{2018}+10^{2017}+8⋮24\left(đpcm\right)\)

//Phần chứng minh chia hết cho 8 em cũng có thể xử lý như sau :

\(10^{2020}=10^3.10^{2017}=1000.10^{2017}=8.125.10^{2017}⋮8\)

Ba lũy thừa của 10 còn lại em tách tương tự sau đó vẫn làm giống như trên .

_Học tốt_