a) Ta có: \(\overline{ababab}=\overline{ab}\cdot10101\) mà \(10101⋮3\) nên \(10101.\overline{ab}⋮3\Rightarrow\overline{ababab}⋮3\)
b)
\(S=5+5^2+5^3+5^4+...+5^{2004}\)
\(=\left(5+5^4\right)+\left(5^2+5^5\right)+\left(5^3+5^6\right)+\left(5^7+5^{10}\right)+\left(5^8+5^{11}\right)+\left(5^9+5^{12}\right)+...+\left(5^{1999}+5^{2002}\right)+\left(5^{2000}+5^{2003}\right)+\left(5^{2001}+5^{2004}\right)\)\(=5.\left(1+5^3\right)+5^2.\left(1+5^3\right)+5^3.\left(1+5^3\right)+5^7.\left(1+5^3\right)+5^8.\left(1+5^3\right)+5^9.\left(1+5^3\right)+...+5^{1999}.\left(1+5^3\right)+5^{2000}.\left(1+5^3\right)+5^{2001}.\left(1+5^3\right)\)\(=\left(1+5^3\right).\left(5+5^2+5^3+...+5^{1999}+5^{2000}+5^{2001}\right)\)
\(=126.\left(5+5^2+5^3+...+5^{1999}+5^{2000}+5^{2001}\right)⋮126\)
Vậy \(S⋮126\)
\(S=5+5^2+5^3+5^4+...+5^{2004}\)
\(=\left(5+5^3\right)+\left(5^2+5^4\right)+\left(5^5+5^7\right)+\left(5^6+5^8\right)+...+\left(5^{2002}+5^{2004}\right)\)\(=5.\left(1+5^2\right)+5^2.\left(1+5^2\right)+5^5.\left(1+5^2\right)+...+5^{2002}.\left(1+5^2\right)\)\(=5.26+5^2.26+5^5.26+...+5^{2002}.26\)
\(=26.\left(5+5^2+5^5+...+5^{2002}\right)\)
\(=26.5.\left(1+5+5^4+5^5+...+5^{2001}\right)\)
\(=130.\left(1+5+5^4+...+5^{2001}\right)⋮65\)
Vậy \(S⋮65\)
