Lời giải:

$A=1+5+5^2+5^3+...+5^{98}+5^{99}$

$=1+(5+5^2+5^3)+(5^4+5^5+5^6)+...+(5^{97}+5^{98}+5^{99})$

$=1+5(1+5+5^2)+5^4(1+5+5^2)+...+5^{97}(1+5+5^2)$

$=1+(1+5+5^2)(5+5^4+...+5^{97})$

$=1+31(5+5^4+....+5^{97})$

$\Rightarrow A$ chia $31$ dư $1$

giúp mình với 

đang cần gấp

A = 3+ 3^3 + 3^5 + 3^7 + ... + 3^97 + 3^99

A=(3+3^3)+(3^5+3^7)+.......+(3^97+3^99)

=30+3^5.(3+3^3)+........+3^97.(3+3^3)

=30+3^5.30+......+3^97.30

\(\Rightarrow\)\(A⋮30\)(Vì các số hạng của tổng \(⋮\)30)

hok tốt!

ta co:3¹+3³=3+27=30\(⋮30\)

A=3¹+3³+3⁵+3⁷+....+3⁹⁷+3⁹⁹

A=1(3+27)+3⁴(3¹+3³)+....+3⁹⁶(3¹+3³)

A=1*30+3⁴*30+...+3⁹⁶*30

A=30* (1+3⁴+...+3⁹⁶)\(⋮30\)

vay \(A⋮30\)

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 \(B⋮41\)

Ai giúp tui đuy mà😭

\(A=\left(1+3+3^2\right)+...+3^{96}\left(1+3+3^2\right)\)

\(=13\cdot\left(1+...+3^{96}\right)⋮13\)