\(B=3^0+3^1+3^2...+3^{100}\)

\(=3^0\times\left(1+3^1+3^2\right)+3^3\times\left(1+3^1+3^2\right)+...+3^{98}\times\left(1+3^1+3^2\right)\)

\(=3^0\times13+3^3\times13+...+3^{98}\times13\)

\(=13\times\left(3^0+3^3+...+3^{98}\right)⋮13\)

B=30+31+32...+3100B=30+31+32...+3100

=30×(1+31+32)+33×(1+31+32)+...+398×(1+31+32)=30×(1+31+32)+33×(1+31+32)+...+398×(1+31+32)

=30×13+33×13+...+398×13=30×13+33×13+...+398×13

=13×(30+33+...+3

Đặt A = 3¹ + 3² + 3³ + 3⁴ + ... + 3⁹⁹ + 3¹⁰⁰

= (3¹ + 3²) + (3³ + 3⁴) + ... + (3⁹⁹ + 3¹⁰⁰)

= 3.(1 + 3) + 3³.(1 + 3) + ... + 3⁹⁹.(1 + 3)

= 3.4 + 3³.4 + ... + 3⁹⁹.4

= 4.(3 + 3³ + ... + 3⁹⁹) ⋮ 4

Vậy A ⋮ 4

.

\(\begin{array}{l}a)M = {32^{2023}} - {32^{2021}}\\M = {32^{2021}}\left( {{{32}^2} - 1} \right)\\M = {32^{2021}}.1023\end{array}\)

Vì \(1023 \vdots 31\) nên \(M = \left( {{{32}^{2021}}.1023} \right) \vdots 31\)

Vậy M chia hết cho 31.

\(\begin{array}{l}b)N = {7^6} + {2.7^3} + {8^{2022}} + 1\\N = {\left( {{7^3}} \right)^2} + {2.7^3} + 1 + {8^{2022}}\\N = {\left( {{7^3} + 1} \right)^2} + {8^{2022}}\\N = {\left( {344} \right)^2} + {8^{2022}}\\N = {\left( {8.43} \right)^2} + {8^{2022}}\\N = {8^2}\left( {{{43}^2} + {8^{2020}}} \right)\end{array}\)

Vì \({8^2} \vdots 8\) suy ra \(N = {8^2}\left( {{{43}^2} + {8^{2020}}} \right) \vdots 8\)

Vậy N chia hết cho 8

\(A=\left(3+3^2+3^3\right)+...+\left(3^{58}+3^{59}+3^{60}\right)\\ A=3\left(1+3+3^2\right)+...+3^{58}\left(1+3+3^2\right)\\ A=\left(1+3+3^2\right)\left(3+...+3^{58}\right)\\ A=13\left(3+...+3^{58}\right)⋮13\)

\(M=\left(2+2^2+2^3+2^4\right)+...+\left(2^{17}+2^{18}+2^{19}+2^{20}\right)\\ M=\left(2+2^2+2^3+2^4\right)+...+2^{16}\left(2+2^2+2^3+2^4\right)\\ M=\left(2+2^2+2^3+2^4\right)\left(1+...+2^{16}\right)\\ M=30\left(1+...+2^{16}\right)⋮5\)