Question

Giải giúp mik vs

Answer 1

\(S=5+5^2+5^3+.............+5^{2004}\)(\(2004\) số hạng)

\(\Leftrightarrow S=\left(5+5^3\right)+\left(5^2+5^4\right)+..........+\left(5^{2002}+5^{2004}\right)\) (\(1007\) nhóm)

\(\Leftrightarrow S=1\left(5+5^3\right)+5^2\left(5+5^3\right)+..........+5^{2002}\left(5+5^3\right)\)

\(\Leftrightarrow S=130+5^2.130+..............+5^{2002}.130\)

\(\Leftrightarrow S=130\left(1+5^2+...........+5^{2002}\right)\)

\(\Leftrightarrow S=65.2\left(1+5^2+..........+5^{2002}\right)⋮65\)

\(\Leftrightarrow S⋮65\rightarrowđpcm\)

Answer 2

\(S=5+5^2+5^3+...+5^{2004}\\ =5\cdot\left(1+5+5^2+...+5^{2003}\right)⋮5\)

\(S=5+5^2+5^3+...+5^{2004}\\ =\left(5+5^2+5^3+5^4\right)+\left(5^5+5^6+5^7+5^8\right)+...+\left(5^{2001}+5^{2002}+5^{2003}+5^{2004}\right)\\ =5\cdot\left(1+5+5^2+5^3\right)+5^5\cdot\left(1+5+5^2+5^3\right)+...+5^{2001}\cdot\left(1+5+5^2+5^3\right)\\ =\left(5+5^5+...+5^{2001}\right)\cdot\left(1+5+5^2+5^3\right)\\ =\left(5+5^5+...+5^{2001}\right)\cdot156\\ =\left(5+5^5+...+5^{2001}\right)\cdot12\cdot13⋮13\)

Vì 5 và 13 là hai số nguyên tố cùng nhau mà \(S⋮5;S⋮13\Rightarrow S⋮5\cdot13\Leftrightarrow S⋮65\)

Vậy \(S⋮65\)

Answer 3

Lời giải:

Ta có: \(S=5+5^2+5^3+....+5^{2004}\)

\(\Rightarrow 5S=5^2+5^3+....+5^{2004}+5^{2005}\)

Do đó \(4S=5^{2005}-5\Rightarrow S=\frac{5^{2005}-5}{4}\)

Ta thấy \(S=\frac{5(5^{2004}-1)}{4}\) mà \(\text{UCLN}(4,5)=1\Rightarrow S\vdots 5\) \((1)\)

Để ý rằng \(5^2\equiv -1\pmod {13}\Rightarrow 5^{2.1002}\equiv 1\pmod {13}\)

Do đó \(5^{2004}-1\vdots 13,\text{UCLN} (4,13)=1\Rightarrow S\vdots 13\) \((2)\)

Từ \((1),(2)\) mà $5$ và $13$ nguyên tố cùng nhau nên \(S\vdots (5.13=65)\)

Ta có đpcm.