CMR: 3^2015 + 3^100 chia hết cho 13 help me
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A=1/2[(7^4)^2008^2015-(3^4)^88^94]
A=1/2.[(...1)-(...1)]
A=1/2.(...0) ma (...0) chia het cho 5 nen 1/2.(...0) chia het cho 5
nen A chia het cho 5.
Vay A chia het cho 5
a) \(3^{10}+3^{11}+3^{12}\)
⇔ \(3^{10}\left(1+3+3^2\right)\)
⇔ \(3^{10}.13\)
⇒ \(3^{10}.13\) chia hết cho 13
Ta có:
\(A=3+3^3+3^5+...+3^{1991}=\left(3+3^3+3^5\right)+\left(3^7+3^9+3^{11}\right)+\left(3^{1987}+3^{1989}+3^{1991}\right)\)
\(A=3.\left(1+3^2+3^4\right)+3^7.\left(1+3^2+3^4\right)+...+3^{1987}.\left(3^{1987}+3^{1989}+3^{1991}\right)\)
\(A=3.91+3^7.91+...+3^{1987}.91=3.7.13+3^7.7.13\)
\(A=13.\left(3.7.13+3^7.7+...+3^{1987}.7\right)\)
Vì: \(A=15.\left(2+2^4+...+2^{58}\right)\)nên \(A⋮13\)
Tương tự:
\(A=\left(3+3^3+3^5+3^7\right)+...+\left(3^{1985}+3^{1987}+3^{1989}+3^{1991}\right)\)
\(A=3.\left(1+3^2+3^4\right)+3^7.\left(1+3^2+3^4\right)+...+3^{1987}.\left(1+3^2+3^4+3^6\right)\)
\(A=3.820+...+3^{1985}.820=3.20.41+...+3^{1985}.20.41\)
\(A=41.\left(3.20+...+3^{1985}.20\right)\)nên \(B⋮41\)
:)
(3+3^3+3^5)+...+(3^1987+3^1989+3^1991)
=3x(1+3^2+3^4)+...+3^1987x(1+3^2+3^4)
=3x91+...+3^1987x91
=(3+...+3^1987)x91=(3+...+3^1987)x13x7\(⋮\)13
Vậy A\(⋮\)13
(3+3^3+3^5+3^7)+...+(3^1985+3^1987+3^1989+3^1991)
=3x(1+3^2+3^4+3^6)+...+3^1985x(1+3^2+3^4+3^6)
=3x820+...+3^1985x820
=(3+...+3^1985)x820=(3+...+3^1985)x41x20\(⋮\)41
Vậy A\(⋮\)41
\(3^3\equiv1\left(mod13\right)\)
\(\Rightarrow3^{99}\equiv1\left(mod13\right)\)
\(\Rightarrow3^{100}\equiv3\left(mod13\right)\)
\(\Rightarrow3^{100}-3\equiv0\left(mod13\right)\)
\(\Rightarrow3^{100}-3⋮13\)
Lời giải:
$3^{2015}+3^{100}=3^{100}(3^{1915}+1)$
Ta thấy:
$3^3\equiv 1\pmod {13}$
$\Rightarrow 3^{1915}=(3^3)^{638}.3\equiv 1^{638}.3\equiv 3\pmod {13}$
$\Rightarrow 3^{1915}+1\equiv 4\pmod {13}$
Ta thấy: $3^{100}\not\vdots 13; 3^{1915}+1\not\vdots 13$
$\Rightarrow 3^{100}(3^{1915}+1)\not\vdots 13$
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