Ta có :

\(5^{70}=\left(5^2\right)^{35}=25^{35}=\left(12.2+1\right)^{35}\equiv1\left(mod12\right)\)

\(7^{70}=\left(7^2\right)^{35}=49^{35}=\left(12.4+1\right)^{35}\equiv1\left(mod12\right)\)

\(\Rightarrow5^{70}+7^{50}\equiv2\left(mod12\right)\) hay \(5^{70}+7^{50}\) chia 12 dư 2

\(5^2\equiv1\left(mod12\right)\Rightarrow5^{2010}\equiv1\left(mod12\right)< 1>.\)

\(7^2\equiv1\left(mod12\right)\Rightarrow7^{10}\equiv1\left(mod12\right)< 2>.\)

\(Từ< 1>và< 2>\Rightarrow5^{2010}+7^{10}\equiv2\left(mod12\right).\)

\(\Rightarrow5^{2010}+7^{10}:12dư2.\)

Vậy \(5^{2010}+7^{10}:12dư2\)

dư 0 duyệt đi

\(\text{Giải}\)

\(5^{70}+7^{50}=25^{35}+49^{25}\)

\(25\equiv1\left(\text{mod 12}\right);49\equiv1\left(\text{mod 12}\right)\)

\(\Rightarrow5^{70}+7^{50}\equiv\left(1+1\right)\left(\text{mod 12}\right)\equiv2\left(\text{mod 12}\right)\)

\(\Rightarrow\text{5^70+7^50 chia 12 dư 2}\)

ta có : \(5^2\equiv1\)( mod 12 ) \(\Rightarrow\left(5^2\right)^{35}\equiv1\)( mod 12 )

hay \(5^{70}\equiv1\)( mod 12 )  (1)  

 \(\Rightarrow\left(7^2\right)\equiv1\)( mod 12 ) \(\Rightarrow\left(7^2\right)^{25}\equiv1\)( mod 12 ) hay \(7^{50}\equiv1\)( mod 12 ) ( 2 )

từ ( 1 ) ; ( 2 )  suy ra \(5^{70}+7^{50}\div12\) dư 2

bài này tớ lm r

cậu ko nên copy bài >:

1, Dễ thấy : \(5^2=25\equiv1\left(mod12\right)\)                                         \(7^2=49\equiv1\left(mod12\right)\)

             \(\rightarrow\left(5^2\right)^{35}\equiv1^{35}\left(mod12\right)\)                                     \(\rightarrow\left(7^2\right)^{35}\equiv1^{35}\left(mod12\right)\)

           \(\rightarrow5^{70}\equiv1\left(mod12\right)\)                                                 \(\rightarrow7^{70}\equiv1\left(mod12\right)\)

Vậy \(5^{70}:12\left(dư1\right)\) và \(7^{70}:12\left(dư1\right)\)Vậy \(\left(5^{70}+7^{70}\right):12\left(dư2\right)\)

Bài 2 :  Ta có : 3012 = 13.231 + 9

Do đó: 3012 đồng dư với 9 (mod13)

=> \(3012^3\)đồng dư với \(9^3\left(mod13\right)\). Mà \(9^3=729\)đồng dư với 1 (mod13)

=> \(3012^3\)đồng dư với 1 (mod13)

Hay \(3012^{93}\)đồng dư với 1 (mod13)

=> \(3012^{93}-1\)đồng dư với 0 (mod13)

Hay \(3012^{93}-1⋮13\left(đpcm\right)\)

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