Mình nghĩ đề bài phải là chia hết cho 59 chứ bạn.

Ta có:

\(10^6-5^7\)

\(=\left(2.5\right)^6-5^7\)

\(=2^6.5^6-5^7\)

\(=2^6.5^6-5^6.5\)

\(=5^6.\left(2^6-5\right)\)

\(=5^6.59\)

Vì \(59⋮59\) nên \(5^6.59⋮59\)

=> \(10^6-5^7⋮59\left(đpcm\right).\)

Chúc bạn học tốt!

a) = 5³. 5²- 5³ .5+ 5³

= 5³ .( 5²- 5+1)

=5^3. 21 mà 21 chia hết cho 7

=) 5⁵ - 5⁴ + 5³ chia hết cho 7

b)= 7⁴.7² + 7⁴.7 -7⁴

= 7⁴( 7²+ 7-1)

=7⁴. 55 mà 55chia hết cho 11

=)7^6 + 7⁵-7⁴ chia hết cho 11

c)=( 2.3.4)^2.27 . (2.27)^2.3.4 . ( 2)^2.5

= ( 6. 4) ^6.9 . ( 6. 9 ) ^6.4. 2¹⁰

⁼ 24⁶. 24⁹. 54⁶.54⁹ . 2¹⁰

⁼1296⁶ . 1296⁴.2¹⁰mà 1296 chia hết cho 72 ( vì 1296 : 72 bằng 18)

=)24^54. 54^24 + 2^10 chia hết cho 72 ^53

ê ko tick à

a)

Ta có :

10⁶ + 5⁷ 

= (2 x 5)⁶ + 5⁷ 

= 2⁶ x 5⁶ + 5⁷ 

= 2⁶ x 5⁶ + 5⁶ x 5

= 5⁶ x (2⁶ + 5)

= 5⁶ x  69

Vì 69 ⋮ 69 => 5⁶ ⋮ 69 => 10⁶ + 5⁷ ⋮ 69

b)

Ta có :

2²⁰ - 2¹⁷ 

= 2¹⁷ x 2³ - 2¹⁷ x 1

= 2¹⁷ x (2³ - 1)

= 2¹⁷ x 7

Vì 7 ⋮ 7 => 2¹⁷ x 7 ⋮ 7 => 2²⁰ - 2¹⁷ ⋮ 7

k nha bn !!!

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

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

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 \(C⋮13\)

#\(Toru\)

a, ta có
C = 5 + 5^2 + 5^3 + 5^4 + ... + 5^20
=> C = 5 . ( 1 + 5 + 5^2 + 5^3 + ... + 5^19 )
=> C chia hết cho 5
b,
C = 5 + 5^2 + 5^3 + 5^4 + ... + 5^20
=> C = 5 . ( 1 + 5 ) + 5^3 . ( 1 + 5 ) + ... + 5^19 . ( 1 + 5 )
=> C = 5 . 6 + 5^3 . 6 + ... + 5^19 . 6
=> C = 6 . ( 5 + 5^3 + ... + 5^19 )
=> C chia hết cho 6
c,
C = 5 + 5^2 + 5^3 + ... + 5^20
=> C = (5 + 5^2 + 5^3 + 5^4 ) + ... + (5^17 + 5^18 + 5^19 + 5^20 )
=> C = 5 . ( 1 + 5 + 5^2 + 5^3 ) + ... + 5^17 . ( 1+ 5 + 5^2 +5^3)
=> C = 5 . 156 + 5^5 . 156 + ...+ 5^17 . 156
=> C = 5 . 12 . 13 + 5^5 . 12 . 13 + ... + 5^17 . 12 . 13
=> C = 13 . ( 5 . 12 + 5^5 . 12 + ... + 5^17 . 12 )
=> C chia hết cho 13

Bài 5: 

b: Ta có: \(n+6⋮n+2\)

\(\Leftrightarrow n+2\in\left\{2;4\right\}\)

hay \(n\in\left\{0;2\right\}\)

c: Ta có: \(3n+1⋮n-2\)

\(\Leftrightarrow n-2\in\left\{-1;1;7\right\}\)

hay \(n\in\left\{1;3;9\right\}\)

7^6+7^5-7^4=7^4*(7^2+7-2)=7^4*55=7^4*5*11 chia hết cho 11

10^9+10^8+10^7=10^7*(10^2+10+1)=10^7*111=10^6*5*222 chi hết cho 222

Bạn có chăc chắn đúng k

chắc chắn luôn

ta có7⁶+7⁵+7⁴=7⁴x(7²+7-1)

                       =7⁴x55

do 55 chia hết cho 11 nên 7⁴x55 chia hết cho 11

vậy7⁶+7⁵-7⁴ chia hết cho 11

a)7⁶+7⁵-7⁴

=7⁴(7²+7-1)

=7⁴.55

Vì 55 chia hết cho 11 nên 7⁴.55 chia hết cho 11

hay 7⁶+7⁵-7⁴ chia hết ch0 11.

b)10⁹+10⁸+10⁷

=10⁷(10²+10+1)

=10⁷.111

=10⁶.10.111

=10⁶.1110

Vì 1110 chia hết cho 222 nên ...

...

Sửa câu a

a)Ta có:

\(A=3+3^2+3^3+...+3^{99}\)

 \(A=\left(3+3^2+3^3\right)+...+\left(3^{97}+3^{98}+3^{99}\right)\) 

\(A=\left(3+3^2+3^3\right)+...+3^{96}.\left(3+3^2+3^3\right)\)

\(A=39+...+3^{96}.39\)

\(A=39.\left(1+...+3^{96}\right)\)

Vì 39 \(⋮\) 13 nên 39 . ( 1 + ... + 3⁹⁶ ) \(⋮\) 13

Vậy A \(⋮\) 13

_________

b)Ta có:

 \(B=5+5^2+5^3+...+5^{50}\)

\(B=\left(5+5^2\right)+\left(5^3+5^4\right)+...+\left(5^{49}+5^{50}\right)\)

\(B=\left(5+5^2\right)+5^2.\left(5+5^2\right)+...+5^{48}.\left(5+5^2\right)\)

\(B=30+5^2.30+...+5^{48}.30\)

\(B=30.\left(1+5^2+...+5^{48}\right)\)

Vì 30 \(⋮\) 6 nên 30. ( 1 + 5² + ... + 5⁴⁸ ) \(⋮\) 6

Vậy B \(⋮\) 6

a,A=3+3²+3³+..+3⁹⁹=(3+3²+3³)+...+(3⁹⁷+3⁹⁸+3⁹⁹)

     =3(1+3+3²)+...+3⁹⁷(1+3+3²)=3x13+...+3⁹⁷x13=13(3+...+97)⋮13

b,B=5+5²+...+5⁵⁰=(5+5²)+...+(5⁴⁹+5⁵⁰)=5(1+5)+..+5⁴⁹(1+5)

  =5x6+...+5⁴⁹x6=6(5+..+5⁴⁹)⋮6.

Bài 1:

\(a,A=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{2009}+2^{2010}\right)\\ A=\left(1+2\right)\left(2+2^3+...+2^{2009}\right)=3\left(2+...+2^{2009}\right)⋮3\\ A=\left(2+2^2+2^3\right)+...+\left(2^{2008}+2^{2009}+2^{2010}\right)\\ A=\left(1+2+2^2\right)\left(2+...+2^{2008}\right)=7\left(2+...+2^{2008}\right)⋮7\)

\(b,\left(\text{sửa lại đề}\right)B=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{2009}+3^{2010}\right)\\ B=\left(1+3\right)\left(3+3^3+...+3^{2009}\right)=4\left(3+3^3+...+3^{2009}\right)⋮4\\ B=\left(3+3^2+3^3\right)+...+\left(3^{2008}+3^{2009}+3^{2010}\right)\\ B=\left(1+3+3^2\right)\left(3+...+3^{2008}\right)=13\left(3+...+3^{2008}\right)⋮13\)

Bài 2:

\(a,\Rightarrow2A=2+2^2+...+2^{2012}\\ \Rightarrow2A-A=2+2^2+...+2^{2012}-1-2-2^2-...-2^{2011}\\ \Rightarrow A=2^{2012}-1>2^{2011}-1=B\\ b,A=\left(2020-1\right)\left(2020+1\right)=2020^2-2020+2020-1=2020^2-1< B\)

đúng rùi