Ta có: 2¹²+1=2⁴.2⁸+1=(2⁴+1).2⁸=17.2⁸

⇒2¹²+1⋮17

\(2^{12}+1\)

\(=\left(2^4+1\right)\left(2^8-2^4+1\right)\)

\(=17\cdot241⋮17\)

Bài 1:

 ta có 3^3 = 27 chia 13 dư 1

=> (3^3)^670 = 3^ 2010 chia 13 dư 1 (1) 
5^2 = 25 chia 13 dư (-1)

=> (5^2)^1005 chia 13 dư (-1)^ 1005 = (-1) (2) 
Từ (1); (2)

=> 3^2010+5^2010 chia 13 dư 1 + (-1) = 0 
hay 3^2010+5^2010 chia hết cho 13. 

bài 1:

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52010=(52)1005≡(−1)1005(mod13)" role="presentation" style="border:0px; color:rgb(40, 40, 40); direction:ltr; display:inline-table; float:none; font-family:helvea,arial,sans-serif; font-size:16.38px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:1px 0px; position:relative; white-space:nowrap" class="MathJax_CHTML mjx-chtml">52010=(52)1005≡(−1)1005(mod13)
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\(A=1+4+4^2+...+4^{99}\)

\(A=\left(1+4+4^2+4^3\right)+\left(4^4+4^5+4^6+4^7\right)+...+\left(4^{96}+4^{97}+4^{98}+4^{99}\right)\)

\(A=85+4^7\left(1+4+4^2+4^3\right)...+4^{96}\left(1+4+4^2+4^3\right)\)

\(A=85+4^7.85+...+4^{96}.85\)

\(A=85.\left(1+4^7+...+4^{96}\right)\)

Vì 85 chia hết cho 17 nên A chia hết cho 17

Sửa đề: \(11\cdot5^{2n}+2^{3n+2}+2^{3n+1}\)

Ta có: \(11\cdot5^{2n}+2^{3n+2}+2^{3n+1}\)

\(=11\cdot25^n+8^n\cdot4+8^n\cdot2\)

\(=11\cdot25^n+6\cdot8^n\)

Vì \(25\equiv8\)(mod 17)

nên \(11\cdot25^n+6\cdot8^n\equiv11\cdot8^n+6\cdot8^n\equiv17\cdot8^n\equiv0\)(mod 17)

hay \(11\cdot5^{2n}+2^{3n+2}+2^{3n+1}⋮17\)(đpcm)

\(2^{12}+1=\left(2^4+1\right)\left(2^8-2^4+1\right)=17\cdot\left(2^8-2^4+1\right)⋮17\)