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S=1+5^2+5^3+...+5^2010
S=1+(5^1+5^2)+...+(5^2009+5^2010)
S=1+5(1+5)+5^3(1+5)+...+5^2009(1+5)
S=1+5.6+5^3.6+...+5^2009.6
S=1+6(5+5^3+5^5+...+5^2009)
Ta có 6(5+5^3+...+5^2009) chia hết cho 2 nên S chia 2 dư 1
S=1+6(5+...+5^2009)=1+6.5(1+5^2+5^4+...+5^2008)
S=1+30(5^2+...+5^2008)
Ta có 30(1+5^2+...+5^2008) chia hết cho 10 nên S chia 10 dư 1
![](https://rs.olm.vn/images/avt/0.png?1311)
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\(S=1+5+5^2+5^3+5^4+...+5^{2010}\)
\(\Leftrightarrow S=1+\left(5+5^2\right)+\left(5^3+5^4\right)+....+\left(5^{2009}+5^{2010}\right)\)
\(\Leftrightarrow S=1+5\left(1+5\right)+5^3\left(1+5\right)+....+5^{2009}\left(1+5\right)\)
\(\Leftrightarrow S=1+5\cdot6+5^3\cdot6+...+5^{2009}\cdot6\)
\(\Leftrightarrow S=1+6\left(5+5^3+...+5^{2009}\right)\)
Mà \(6\left(5+5^3+....+5^{2009}\right)⋮2\)
=> S chia 2 dư 1
![](https://rs.olm.vn/images/avt/0.png?1311)
S=17+5^2(1+5+5^2)+...+5^2008(1+5+5^2)
=17+5^2.31+...+5^2008.31
=17+ 31(5^2+...+5^2008) chia cho 31 dư 17
Nhớ k mình nhé!!!!!