\(2^5=32\equiv1\left(mod31\right)\)

\(\Rightarrow\left(2^5\right)^{400}\equiv1\)( mod 31)

\(\Rightarrow2^{2000}\equiv1\)( mod 31)

\(\Rightarrow2^{2000}\times2^2\equiv2^2\)( mod 31)

\(\Rightarrow2^{2002}\equiv4\)( mod 31)

\(\Rightarrow2^{2002}-4\equiv0\)( mod 31)

iwjdfìewaohdòihódfuhtAao xdem sssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssex lko dSVOKJDưgeohqởigie

2⁵ = 32 = 1 (mod 31)

=> (2⁵)⁴⁰⁰ = 1⁴⁰⁰ = 1 (mod 31)

=> 2²⁰⁰⁰ = 1 (mod 31)

=> 2²⁰⁰⁰.2² = 2² (mod 31)

=> 2²⁰⁰² = 4 (mod 31)

=> 2²⁰⁰² - 4 = 0 (mod 31)

Vậy... 

Bạn vào câu hỏi tương tự nhé !!!

  2222 ≡ 3 (mod 7) ; 3³ ≡ -1 (mod 7) ; chú ý: 5555 = 3*1851 + 2 
=> 2222^5555 ≡ 3^5555 ≡ (3³)^1851.3² ≡ (-1)^1851.9 ≡ -9 ≡ -2 ≡ 5 (mod 7) 

5555 ≡ 4 (mod 7) ; 4³ ≡ 1 (mod 7) ; 2222 = 3*740 + 2 
=> 5555^2222 ≡ 4^2222 ≡ (4³)^740.4² ≡ (1).16 ≡ 2 (mod 7) 

vậy: 2222^5555 + 5555^2222 ≡ 5+2 ≡ 0 (mod 7) => đpcm 

\(2^{1995}-1=A=1+2+2^2+2^3+2^4...+2^{1994}\)

\(\left(1+2+2^2+2^3+2^4\right)=31\) chia hết cho 31

Số số hạng của A là 1995 chia hết cho 5 

\(A=31.\left(1+2^5+2^{10}+..+2^{\frac{1995}{5}-5}\right)\)=> DPCM

\(B=3+3^2+3^3+3^4+...+3^{2009}+3^{2010}\)

\(=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{2009}+3^{2010}\right)\)

\(=3\left(1+3\right)+3^3\left(1+3\right)+...+3^{2009}\left(1+3\right)\)

\(=4.\left(3+3^3+...+3^{2009}\right)\)

⇒ \(B\) ⋮ 4

b: \(C=5\left(1+5+5^2\right)+...+5^{2008}\left(1+5+5^2\right)=31\cdot\left(5+...+5^{2008}\right)⋮31\)

2^1995 - 1 = ( 2^5)^399 = 32^399 -1

Ma 32 dong du vs 1( mod 31 )

=> 32^399 dong du vs 1( mod 31 )

=> 32^399 dong du vs 0( mod 31 )

=> 2^1995 - 1 chia het cho 31 ( dpcm ) 

Ta có: \(2^{1995}=\left(2^5\right)^{399}=32^{399}⋮32\)

Mà \(32\equiv1\)(mod 31)

\(\Rightarrow2^{1995}\equiv1\)(mod 31)

\(\Rightarrow2^{1995}-1⋮31\)(đpcm)

Ta có : \(2^{1995}=2^{1990}\cdot2^5=2^{1990}\cdot32\)

Vì \(32\div31\)dư 1 \(\Rightarrow32\cdot2^{1990}⋮31\)

vạy \(2^{1995}-1⋮31\)