\(\overline{3ab3}-\overline{3ba3}=3003-3003+\overline{ab0}-\overline{ba0}=10\left(\overline{ab}-\overline{ba}\right)=10\left(10a+b-10b-a\right)\)

\(=10\left(9a-9b\right)=90\left(a-b\right)⋮90\)

\(\overline{aaa}+\overline{bbb}=111.a+111.b=111\left(a+b\right)=37\times3\times\left(a+b\right)⋮37\)

\(\overline{abc}+\overline{bca}+\overline{cab}=\left(100a+10b+c\right)+\left(100b+10c+a\right)+\left(100c+10a+b\right)\)

\(=111\left(a+b+c\right)=37\times3\times\left(a+b+c\right)⋮37\)

\(S=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{11}\left(1+2\right)=3\left(2+2^3+...+2^{11}\right)⋮3\)

\(S=2\left(1+2+2^2\right)+...+2^{10}\left(1+2+2^2\right)=7\left(2+...+2^{10}\right)⋮7\)

Vì S chia hết cho 2 và S chia hết cho 3 

nên \(S⋮6\)

5⁵-5⁴+5³=5³.(5²-5¹+5⁰)=5³.(25-5+1)=5³.21=5³.3.7 chia hết cho 7

=>ĐPCM

7⁶+7⁵-7⁴=7⁴.(7²+7¹-7⁰)=7⁴.(49+7-1)=7⁴.55=7⁴.5.11 chia hết cho 11

=>ĐPCM

1) 5⁵ - 5⁴ + 5³ = 5³ . 5² - 5³ . 5 - 5³

= 5³ . ( 5² - 5 + 1 )

= 5³ . ( 25 - 5 - 1 )

= 5³ . 21

= 5³ . 3 . 7 chia hết cho 7

Vậy chứng minh 5⁵ - 5⁴ + 5³ chia hết cho7

2) 7⁶ + 7⁵ - 7⁴ = 7⁴ . 7² + 7⁴ . 7 - 7⁴

= 7⁴ . ( 7² + 7 - 1 )

= 7⁴ . ( 49 + 7 - 1 )

= 7⁴ . 55

= 7⁴ . 5 .11 chia hết cho 11

Vậy chứng minh 7⁶ + 7⁵ - 7⁴ chia hết cho 11

Tích mình nha !!!!!!!!!!!!!!!!!