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a) Có: \(2^3=8\equiv1\left(mod7\right)\Rightarrow2^{51}\equiv1\left(mod7\right)\)
\(\Rightarrow2^{51}-1⋮7\left(đpcm\right)\)
b) 270 + 370 = (22)35 + (32)35 = 435 + 935
\(=\left(4+9\right).\left(4^{34}-4^{33}.9+....-4.9^{33}+9^{34}\right)\)
\(=13.\left(4^{34}-4^{33}.9+...-4.9^{33}+9^{34}\right)⋮13\left(đpcm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
phần a sai đề nha bạn
b,Ta có
\(2\equiv2\left(mod13\right)\)
\(\Rightarrow2^{12}\equiv1\left(mod13\right)\)
\(\Rightarrow2^{12.5}.2^{10}\equiv1.2^{10}\left(mod13\right)\)
\(\Rightarrow2^{60}.2^{10}\equiv1024\left(mod13\right)\)
\(\Rightarrow2^{70}\equiv10\left(mod13\right)\)\(\left(1\right)\)
Lại có:
\(3\equiv3\left(mod13\right)\)
\(\Rightarrow3^6\equiv1\left(mod13\right)\)
\(\Rightarrow3^{6.11}.3^4\equiv1.3^4\left(mod13\right)\)
\(\Rightarrow3^{66}.3^4\equiv81\left(mod13\right)\)
\(\Rightarrow3^{70}\equiv3\left(mod13\right)\)\(\left(2\right)\)
Từ \(\left(1\right);\left(2\right)\Rightarrow2^{70}+3^{70}\equiv13\equiv0\left(mod13\right)\)
c, Ta có
\(17\equiv-1\left(mod18\right)\)
\(\Rightarrow17^{19}\equiv-1\left(mod18\right)\)\(\left(1\right)\)
Lại có
\(19\equiv1\left(mod18\right)\)
\(\Rightarrow19^{17}\equiv1\left(mod18\right)\)\(\left(2\right)\)
Từ \(\left(1\right);\left(2\right)\Rightarrow17^{19}+19^{17}\equiv0\left(mod18\right)\)
\(\Rightarrow17^{19}+19^{17}⋮18\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1)
a)251-1
=(23)17-1\(⋮\)23-1=7
Vậy 251-1\(⋮\)7
b)270+370
=(22)35+(32)35\(⋮\)22+32=13
Vậy 270+370\(⋮\)13
c)1719+1917
=(BS18-1)19+(BS18+1)17
=BS18-1+BS18+1
=BS18\(⋮\)18
d)3663-1\(⋮\)35\(⋮\)7
Vậy 3663-1\(⋮\)7
3663-1
=3663+1-2
=BS37-2\(⋮̸\)37
Vậy 3663-1\(⋮̸\)37
e)24n-1
=(24)n-1\(⋮\)24-1=15
Vậy 24n-1\(⋮\)15
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có: \(8^5+2^{11}\)
\(=\left(2^3\right)^5+2^{11}\)
\(=2^{15}+2^{11}\)
\(=2^{11}\left(2^4+1\right)\)
\(=2^{11}.17⋮17\left(đpcm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(17^{19}+19^{17}=\left(17^{19}+1\right)+\left(19^{17}-1\right)\)
Mà \(17^{19}+1\)chia hết cho \(17+1=18\)
và \(19^{17}-1\)chia hết cho \(19-1=18\)
nên \(\left(17^{19}+1\right)+\left(19^{17}-1\right)\)chia hết cho \(18\)
Do đó, \(17^{19}+19^{17}\)chia hết cho \(18\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta áp dụng đẳng thức sau: \(a^{2k+1}+b^{2k+1}⋮a+b\)
\(A=2^{70}+b^{70}=4^{35}+9^{35}⋮4+9=13\)
\(\Rightarrowđpcm\)
b) Ta có: \(17\equiv-1\left(mod18\right)\Rightarrow17^{19}\equiv-1\left(mod18\right)\)
\(19\equiv1\left(mod18\right)\Rightarrow19^{17}\equiv1\left(mod18\right)\)
\(\Rightarrow17^{19}+19^{17}⋮18\left(đpcm\right)\)