a: Sửa đề: chia hết cho 10

\(11^{10}-1\)

\(=\left(11-1\right)\left(11^9+11^8+\cdots+11+1\right)\)

\(=10\cdot\left(11^9+11^8+\cdots+11+1\right)\) ⋮10

b: Đặt \(A=9\cdot10^{n}+18\)

\(=9\cdot\left(10^{n}+2\right)\)

\(=9\cdot100\ldots02\)

Tổng các chữ số của số 1000....02 là:

1+0+...+0+2=3

=>\(100\ldots02\) ⋮3

=>\(9\cdot100\ldots02\) ⋮3*9

=>A⋮27

\(a,C=5+5^2+5^3+5^4+\cdot\cdot\cdot+5^{20}\)

\(=5\left(1+5+5^2+\cdot\cdot\cdot+5^{19}\right)\)

Ta thấy: \(5\left(1+5+5^2+\cdot\cdot\cdot+5^{19}\right)⋮5\)

nên \(C⋮5\)

\(b,C=5+5^2+5^3+5^4\cdot\cdot\cdot+5^{20}\)

\(=\left(5+5^2\right)+\left(5^3+5^4\right)+\cdot\cdot\cdot+\left(5^{19}+5^{20}\right)\)

\(=5\left(1+5\right)+5^3\left(1+5\right)+\cdot\cdot\cdot+5^{19}\left(1+5\right)\)

\(=5\cdot6+5^3\cdot6+\cdot\cdot\cdot+5^{19}\cdot6\)

\(=6\cdot\left(5+5^3+\cdot\cdot\cdot+5^{19}\right)\)

Ta thấy: \(6\cdot\left(5+5^3+\cdot\cdot\cdot+5^{19}\right)⋮6\)

nên \(C⋮6\)

\(c,C=5+5^2+5^3+5^4+\cdot\cdot\cdot+5^{20}\)

\(=\left(5+5^3\right)+\left(5^2+5^4\right)+\cdot\cdot\cdot+\left(5^{17}+5^{19}\right)+\left(5^{18}+5^{20}\right)\)

\(=5\left(1+5^2\right)+5^2\left(1+5^2\right)+\cdot\cdot\cdot+5^{17}\cdot\left(1+5^2\right)+5^{18}\left(1+5^2\right)\)

\(=5\cdot26+5^2\cdot26+\cdot\cdot\cdot+5^{17}\cdot26+5^{18}\cdot26\)

\(=26\cdot\left(5+5^2+\cdot\cdot\cdot+5^{17}+5^{18}\right)\)

Ta thấy: \(26\cdot\left(5+5^2+\cdot\cdot\cdot+5^{17}+5^{18}\right)⋮13\)

nên \(C⋮13\)

#\(Toru\)

b: \(\overline{aaa}=100a+10a+a=111a\) \(=3\cdot37a\) ⋮37

c: \(\overline{aaaaaa}=a\cdot10^5+a\cdot10^4+a\cdot10^3+a\cdot10^2+a\cdot10+a\)

\(=a\left(10^5+10^4+10^3+10^2+10+1\right)\)

\(=a\left\lbrack10^4\left(10+1\right)+10^2\left(10+1\right)+\left(10+1\right)\right\rbrack=a\cdot11\cdot\left(10^4+10^2+1\right)\)

\(=a\cdot11\cdot10101=a\cdot11\cdot37\cdot273\) ⋮37

d: \(\overline{abcabc}\)

\(=a\cdot10^5+b\cdot10^4+c\cdot10^3+a\cdot10^2+b\cdot10+c\)

\(=a\cdot10^2\left(10^3+1\right)+b\cdot10\cdot\left(10^3+1\right)+c\left(10^3+1\right)\)

\(=1001\left(100a+10b+c\right)=7\cdot11\cdot13\cdot\left(100a+10b+c\right)\)

=>\(\overline{abcabc}\) chia hết cho cả 13 và 11

1)

a)\(B=3+3^3+3^5+3^7+.....+3^{1991}\)

\(\Leftrightarrow B=3\left(1+3^2+3^4+3^6+.....+3^{1990}\right)\)

Vì \(3\left(1+3^2+3^4+3^6+.....+3^{1990}\right)\)chia hết cho 3 nên \(B⋮3\)

\(B=3+3^3+3^5+3^7+.....+3^{1991}\)

\(\Leftrightarrow B=\left(3+3^3+3^5+3^7\right)+.....+\left(3^{1988}+3^{1989}+3^{1990}+3^{1991}\right)\)

\(\Leftrightarrow B=3\left(1+3^2+3^4+3^6\right)+.....+3^{1988}\left(1+3^2+3^4+3^6\right)\)

\(\Leftrightarrow B=3.820+.....+3^{1988}.820\)

\(\Leftrightarrow B=3.20.41+.....+3^{1988}.20.41\)

Vì \(3.20.41+.....+3^{1988}.20.41\) chia hết cho 41 nên \(B⋮41\)

Lời giải:
a.

\(\overline{abc}=100a+10b+c\)

Vì $a,b$ là số chẵn nên $100a\vdots 4; 10b\vdots b$

Mà $\overline{abc}=100a+10b+c\vdots 4$

$\Rightarrow c\vdots 4$

(đpcm)

b.

$\overline{bac}=100b+10a+c$

$=100a+10b+c+(90b-90a)=\overline{abc}+90(b-a)$

Vì $b,a$ chẵn nên $b-a$ chẵn

$\Rightarrow 90(b-a)=45.2(b-a)\vdots 4$

Kết hợp với $\overline{abc}\vdots 4$

Do đó: $\overline{bac}=\overline{abc}+90(b-a)\vdots 4$

(đpcm)

Tham khảo: https://olm.vn/hoi-dap/detail/67971789293.html

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