Ta có:

\(B=2024\cdot14\)

\(B=2\cdot1012\cdot14\)

\(B=28\cdot1012\)

Vậy B chia hết cho 28

Ta có:

\(C=111111\cdot18\)

\(C=3\cdot7\cdot11\cdot13\cdot37\cdot3^2\cdot2\)

\(C=\left(3\cdot3^2\right)\cdot\left(7\cdot11\cdot13\cdot37\cdot2\right)\)

\(C=3^3\cdot74074\)

\(C=27\cdot74074\)

Vậy C chia hết cho 27

Ta có:

\(11^{2024}\)

\(=11^2\cdot11^{2022}\)

\(=121\cdot11^{2022}\)

Vậy \(11^{2024}\) chia hết cho \(121\)

Ta có:

\(3^{2022}\)

\(=3^2\cdot3^{2020}\)

\(=3^2\cdot3^2\cdot3^{2018}\)

\(=3^4\cdot3^{2018}\)

\(=81\cdot3^{2018}\)

Vậy \(3^{2022}\) chia hết cho 81 

3²⁰²² = 3⁴.3²⁰¹⁸

= 81.3²⁰¹⁸ ⋮ 81

Vậy 3²⁰²² ⋮ 81

Ta có: 

\(A=15\cdot16\)

\(A=3\cdot5\cdot2^4\)

\(A=3\cdot5\cdot2^3\cdot2\)

\(A=2^3\cdot5\cdot3\cdot2\)

\(A=40\cdot6\)

Vậy A chia hết cho 40

a) $A=3+3^2+3^3+3^4+3^5+3^6+....+3^{28}+3^{29}+3^{30}$

$\iff A=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+....+\left(3^{28}+3^{29}+3^{30}\right)$

$\iff A=3\left(1+3+3^2\right)+3^4\left(1+3+3^2\right)+....+3^{28}\left(1+3+3^2\right)$

$\iff A=3.13+3^4.13+....+3^{28}.13$

$\iff A=13\left(3+3^4+....+3^{28}\right)⋮13\left(dpcm\right)$

b) $A=3+3^2+3^3+3^4+3^5+3^6+....+3^{25}+3^{26}+3^{27}+3^{28}+3^{29}+3^{30}$

$\iff A=\left(3+3^2+3^3+3^4+3^5+3^6\right)+....+\left(3^{25}+3^{26}+3^{27}+3^{28}+3^{29}+3^{30}\right)$

$\iff A=3\left(1+3+3^2+3^3+3^4+3^5\right)+....+3^{25}\left(1+3+3^2+3^3+3^4+3^5\right)$

$\iff A=3.364+....+3^{25}.364$

$\iff A=364\left(3+3^5+3^{10}+....+3^{25}\right)$

$\iff A=52.7\left(3+3^5+3^{10}+....+3^{25}\right)⋮52\left(dpcm\right)$

Ta có:2.4.6.8.10=40.2.6.8 chia hết cho 40

          mà 40 chia hết cho 40

nên 2.4.6.8.10+40 chia hết cho 4