Ta có A = \(1+5+5^2+...+5^{2015}\)

=> 5A = \(5+5^2+5^3+...+5^{2016}\)

=> 5A - A =  \(5+5^2+5^3+...+5^{2016}-1-5-5^2-...-5^{2015}\)

=> 4A = \(5^{2016}-1\)

=> A = \(\left(5^{2016}-1\right):4\)

=> A chia hết cho 31

Ta có:

5⁷+5⁸+5⁹

=5⁷(1+5+5²)

=5⁷.31

Vì 31 chia hết cho 31=)5⁷.31 chia hết cho 31

Vậy 5⁷+5⁸+5⁹ chia hết cho 31

Học tốt nhé

c)\(^{5^7+5^8+5^9}\)

= \(5^7\left(1+5+5^2\right)\)

= \(5^7.31\)

\(5^7.31⋮31\)

\(\Rightarrow\)\(5^7+5^8+5^9\)\(⋮\)\(31\)

a/ Ta có :

\(9^{1945}-2^{1930}=\left(9^5\right)^{389}-\left(2^{10}\right)^{193}=\left(.....9\right)-\left(.....4\right)=\left(............5\right)⋮5\)

\(\Leftrightarrowđpcm\)

Tìm 2B

Lấy 2B trừ B

B chia hết cho 30 :
B = 5 + 5² + ... + 5⁹⁶ 

B = ( 5 + 5² ) + ( 5³ + 5⁴ ) + ... + ( 5⁹⁵ + 5⁹⁶ )

B = 5 ( 1 + 5 ) + 5³ ( 1 + 5 ) + ... + 5⁹⁵ ( 1 + 5 )

B = 5 . 6 + 5³ . 6 + ... + 5⁹⁵ . 6
B = 6 ( 5 + 5³ + ... + 5⁹⁵ )

= > B chia hết cho 6
Vì B các số hạng của B là những số chia hết cho 5 ( 5 ; 5² ; ... ; 5⁹⁶ )

= > Tổng B chia hết cho 5
Vì ( 5 ; 6 ) = 1 = > B chia hết cho 30

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khó ghê , bài này đẳng cấp ghê

giả sử a chia hết cho 5

=>a² chia hết cho 5

=>a²-1 không chia hết cho 5

nếu a²-1 chia hết cho 5

=>a² đồng dư với 1(mod 5)

=>a đồng dư với -1 hoặc 1(mod 5)

=>a có tận cùng là 4;6;1;9 

=>đpcm

^-^

\(5^{10}+5^9+5^8=5^8.\left(5^2+5+1\right)=5^8.31\) chia hết cho 31

\(5^{10}+5^9+5^8=5^8\left(5^2+5+1\right)\)

\(=5^8\left(25+5+1\right)=5^8.31⋮31\)

Vậy biểu thức trên chia hết cho 31

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

\(=5^3.21⋮7\)

Ta có: 5⁵ - 5⁴ + 5³

= 5³ (5² - 5 + 1)

= 5³ . 21

Vì 21 chia hết cho 7 nên 5³ . 21 chia hết cho 7

hay 5⁵ - 5⁴ + 5³ chia hết cho 7