a) \(A=4+4^2+4^3+...+4^{60}=4\left(1+4+4^2+...+4^{59}\right)⋮4\)

b) \(A=4+4^2+4^3+...+4^{60}=4\left(1+4\right)+4^3\left(1+4\right)+...+4^{59}\left(1+4\right)=4.5+4^3.5+...+4^{59}.5=5\left(4+4^3+...+4^{59}\right)⋮5\)

c) \(A=4+4^2+4^3+...+4^{60}=4\left(1+4+4^2\right)+4^4\left(1+4+4^2\right)+...+4^{58}\left(1+4+4^2\right)=4.21+4^4.21+...+4^{58}.21=21\left(4+4^4+...+4^{58}\right)⋮21\)

\(A=4+4^2+4^3+.....+4^{60}\)

\(A=\left(1+4+4^2\right)+4^3.\left(1+4+4^2\right)+....4^{57}.\left(1+4+4^2\right)\)

\(A\)\(=21+4^3.21+...4^{57}.21\)

\(\Rightarrow A⋮4;21\)

ko chia hết cho 5

a:Ta có: \(A=4+4^2+4^3+...+4^{60}\)

\(=4\left(1+4+4^2+...+4^{59}\right)⋮4\)

b: Ta có: \(A=4+4^2+4^3+...+4^{60}\)

\(=4\left(1+4\right)+4^3\left(1+4\right)+...+4^{59}\left(1+4\right)\)

\(=5\cdot\left(4+4^3+...+4^{59}\right)⋮5\)

a) S= 2 + 2² + 2³ +...+ 2¹⁰⁰

S= ( 2+2² ) + ( 2³+2⁴ ) +...+( 2⁹⁹ + 2¹⁰⁰ )

S= 6+ 2² ( 2+2²)+ ...+ 2⁹⁸ (2+2²)

S=6+ 2².6+ ...+ 2⁹⁸.6

S= 6.(2²+...+2⁹⁸) chia hết cho 3 ( vì 6 chia hết cho 3)

4 chia 3 là tứ chia tam->tám chia tư=2

Ta có:

4 : 3 = Tứ chia tam = tám chia tư = 8 : 4 = 2

Vậy 4 : 3 = 2

Hok tốt @

Trả lời :

4 : 3 = Tứ : Tam = Tám : Tư = 8 : 4 = 2

Hok tốt

-Ken-

1.Áp dụng định lý Fermat nhỏ.

1) \(a^5-a=a\left(a^4-1\right)=a\left(a^2-1\right)\left(a^2+1\right)\)

\(=\left(a-1\right)a\left(a+1\right)\left(a^2-4+5\right)\)

\(=\left(a-1\right)a\left(a+1\right)\left(a^2-4\right)+5\left(a-1\right)a\left(a+1\right)\)

\(=\left(a-2\right)\left(a-1\right)a\left(a+1\right)\left(a+2\right)+5\left(a-1\right)a\left(a+1\right)⋮5\)

Vì \(\left(a-2\right)\left(a-1\right)a\left(a+1\right)\left(a+2\right)⋮5\)( tích 5 số nguyên liên tiếp chia hết cho 5)

và \(5\left(a-1\right)a\left(a+1\right)⋮5\)

=> \(a^5-a⋮5\)

Nếu \(a^5⋮5\)=> a chia hết cho 5

Cách 2

\(a^5-a=a\left(a^4-1\right)=a\left(a^2-1\right)\left(a^2+1\right)\)

\(=a\left(a-1\right)\left(a+1\right)\left(a^2+1\right)\)

Do a nguyên nên a có 5 dạng:\(5k;5k+1;5k+2;5k+3;5k+4\)

Nếu \(a=5k\Rightarrow a^5-a=5k\left(a-1\right)\left(a+1\right)\left(a^2+1\right)⋮5\)

Nếu \(a=5k+1\Rightarrow a^5-a=a\cdot5k\left(a+1\right)\left(a^2+1\right)⋮5\)

Nếu \(a=5k+2\Rightarrow a^5-a=a\left(a-1\right)\left(a+1\right)\left(25k^2+20k+5\right)⋮5\)

Nếu \(a=5k+3\Rightarrow a^5-a=a\left(a-1\right)\left(a+1\right)\left(25k^2+30k+10\right)⋮5\)

Nếu \(a=5k+4\Rightarrow a^5-a=a\left(a-1\right)\left(5k+5\right)\left(a^2+1\right)⋮5\)

Vậy \(a^5-a⋮5\)

c)D=4+4²+4³+4⁴+...+4²⁰¹²

D=(4+4²)+(4³+4⁴)+...+(4²⁰¹¹+4²⁰¹²)

D=4.5+4³.5+4⁵.5+...+4²⁰¹¹.5

D=5.(4+4³+4²⁰¹¹)

=>D chia hết cho 5

=>ĐPCM

tất cả đều có trong câu hỏi tương tự

b)

A=(1+5+5²)+(5³+5⁴+5⁵)+...(5⁴⁰²+5⁴⁰³+5⁴⁰⁴)

A=31.1+31.5³+...+31.5⁴⁰²

A=31.(1+5³+...+5⁴⁰²)

=>A chia hết cho 31

=>Đâu phải con ma

Minh lam cau A) thoi duoc hong