a, 6¹⁰⁰ - 1 = (6 . 6 . 6 ..... 6) - 1 = [(...6) . (...6) . (...6) ..... (...6)] - 1 = (...6) - 1 = ...5 \(⋮\) 5

b, 21²⁰ - 11¹⁰ = (21 . 21 . 21 . 21 . 21..... 21) - (11 . 11 . 11 . 11 ..... 11) = [(...1) . (...1) . (...1) . (...1).....(...1)] - [(...1) . (...1) . (...1) . (...1).....(...1)] = (...1) - (...1) = ....0 \(⋮\) 2; \(⋮\) 5

\(10^{2120}=1000\ldots0\) (2120 chữ số 0)

Tổng các chữ số của số \(10^{2120}\) là:

1+0+0+...+0=1

=>\(10^{2120}\) chia 3 dư 1(1)

2120=2118+2=3*706+3

=>2120 chia 3 dư 2(2)

Từ (1),(2) suy ra \(10^{2120}+2120\) ⋮3

Ta có: \(10^{2120}\) ⋮10

2120⋮10

Do đó: \(10^{2120}+2120\) ⋮10

mà \(10^{2120}+2120\) ⋮3

và ƯCLN(3;10)=1

nên \(10^{2120}+2120\) ⋮3*10

=>\(10^{2120}+2120\) ⋮30

\(a,\left(n+10\right)\left(n+15\right)\)

Với n lẻ \(\Rightarrow n=2k+1\left(k\in N\right)\)

\(\Rightarrow\left(n+10\right)\left(n+15\right)=\left(2k+11\right)\left(2k+16\right)=2\left(k+8\right)\left(2k+11\right)⋮2\)

Với n chẵn \(\Rightarrow n=2q\left(q\in N\right)\)

\(\Rightarrow\left(n+10\right)\left(n+15\right)=\left(2q+10\right)\left(2q+15\right)=2\left(q+5\right)\left(2q+15\right)⋮2\)

Suy ra đpcm

\(b,\) Với n chẵn \(\Rightarrow n=2k\Rightarrow n\left(n+1\right)\left(2n+1\right)⋮2\)

Với n lẻ \(\Rightarrow n=2q+1\Rightarrow n+1=2q+2=2\left(q+1\right)⋮2\Rightarrow n\left(n+1\right)\left(2n+1\right)⋮2\)

Vậy \(n\left(n+1\right)\left(2n+1\right)⋮2\)

Với \(n=3k\Rightarrow n\left(n+1\right)\left(2n+1\right)⋮3\)

Với \(n=3k+1\Rightarrow2n+1=6k+3=3\left(2k+1\right)⋮3\Rightarrow n\left(n+1\right)\left(2n+1\right)⋮3\)

Với \(n=3k+2\Rightarrow n+1=3\left(k+1\right)⋮3\Rightarrow n\left(n+1\right)\left(2n+1\right)⋮3\)

Vậy \(n\left(n+1\right)\left(2n+1\right)⋮3\)

Suy ra đpcm

a+5b ⋮ 7
=> 3(a+5b) ⋮7
=> 3a+15b⋮7
=> 3a+15b +7a -14b⋮7
=> 10a+b⋮7
chúc bn hok tốt ^_^

Ta có: \(\overline{abc}-\overline{def}\) ⋮13

\(1001\cdot\overline{abc}\) ⋮13(Vì \(1001=77\cdot13\) )

=>\(1001\cdot\overline{abc}-\left(\overline{abc}-\overline{def}\right)\) ⋮13

=>\(\overline{abc}\cdot1000+\overline{def}\) ⋮13

=>\(\overline{abcdef}\) ⋮13

ta có :

A chia hết cho 15 nên A chia hết cho 3 và A chia hết cho 5

a) B\(=\) 3 + 3² + 3³ + ... + 3⁶⁰ 

\(=\)(3+3²)+(3³+3⁴)+...+(3⁵⁹+3⁶⁰)

\(=\)3(1+3)+3³(1+3)+...+3⁵⁹(1+3)

\(=\)(3+1)(3+3³+...+3⁵⁹)

\(=\)4(3+3³+...+3⁵⁹)⋮4

⇒B⋮4

b) B\(=\)(3+3²+3³)+...+(3⁵⁸+3⁵⁹+3⁶⁰)

\(=\)3⁰(3+3²+3³)+...+3⁵⁷(3⁵⁸+3⁵⁹+3⁶⁰)

\(=\)3+3²+3³(3⁰+3³+3⁶+...+3⁵⁷)

\(=\)39(3⁰+3³+3⁶+...+3⁵⁷)⋮13

⇒ B⋮13

a: TH1: 2x+5y⋮13

=>22x+55y⋮13

=>22x+16y+39y⋮13

=>22x+16y⋮13

=>2(11x+8y)⋮13

=>11x+8y⋮13

=>(2x+5y)(11x+8y)⋮13*13

=>A⋮169(1)

TH2: 11x+8y⋮13

=>22x+16y⋮13

=>22x+16y+39y⋮13

=>22x+55y⋮13

=>11(2x+5y)⋮13

=>2x+5y⋮13

=>(11x+8y)(2x+5y)⋮13*13

=>A⋮169(2)

Từ (1),(2) suy ra A⋮169

b: 4x+7y⋮23

=>44x+77y⋮23

=>44x+8y+69y⋮23

=>44x+8y⋮23

=>4(11x+2y)⋮23

=>11x+2y⋮23

c: 3x+12y⋮13

=>30x+120y⋮13

=>30x+3y+117y⋮13

=>30x+3y⋮13

=>3(10x+y)⋮13

=>10x+y⋮13