a) A chia hết cho 2 vì tất cả các số hạng của tổng đều chia hết cho 2.

b) Ta tách ghép các số hạng của A thành các nhóm sao cho mỗi nhóm xuất hiện thừa số chia hết cho 3. Khi đó:

4₁ A= 2 +2²³ +2 ² + + 220 a₁ A = 2₁ [1 + 2 +2²¹ +. +2¹2):2 Vay A chia hết choi b₁ A = 2 + 2² +2²+ + 220 (2 +2²) + (2 ² + 2 9) + . + (219+220) = 2₁ (1 + 2) + 2² (2+1). .. +2 19 (2+1) + = 2₁3 + 2³.3 + ..+ 219.3. = (2+2 ³+ + 219) 3:3 Vậy A chia hết cho 3 A = 2 + 2 ² + 2³ + 2ª +. 20 + 2.9+ +2 2+2 ³ + 2² +2²4 + + 218 + 720 +2²³ +2²+ +218 +220 2. (2 +2²) + 2² (1+2²) +.. + 218 ( 1 +2²) = 2 5 +2²5 + + 218 5. 12 +2° + 2 ... +218 ) 5 : 5. vậy A chia hết cho 5

vì x chia hết cho 23 và 11 nhân y cũng chia hết cho 23

`#3107.101107`

a,

\(C=2+2^3+2^5+...+2^{23}\)

\(=\left(2+2^3+2^5\right)+\left(2^5+2^7+2^9\right)+...+\left(2^{19}+2^{21}+2^{23}\right)\)

\(=2\left(1+2^2+2^4\right)+2^5\cdot\left(1+2^2+2^4\right)+...+2^{19}\cdot\left(1+2^2+2^4\right)\)

\(=\left(1+2^2+2^4\right)\cdot\left(2+2^5+...+2^{19}\right)\)

\(=21\cdot\left(2+2^5+...+2^{19}\right)\)

Vì \(21\text{ }⋮\text{ }21\)

\(\Rightarrow21\left(2+2^5+...+2^{19}\right)\text{ }⋮\text{ }21\)

Vậy, \(C\text{ }⋮\text{ }21\)

b,

\(C=2+2^3+2^5+...+2^{23}\)

\(=\left(2+2^3\right)+\left(2^5+2^7\right)+...+\left(2^{21}+2^{23}\right)\)

\(=\left(2+2^3\right)+2^4\cdot\left(2+2^3\right)+...+2^{20}\cdot\left(2+2^3\right)\)

\(=\left(2+2^3\right)\cdot\left(1+2^4+...+2^{20}\right)\)

\(=10\cdot\left(1+2^4+...+2^{20}\right)\)

Vì \(10\text{ }⋮\text{ }10\)

\(\Rightarrow10\cdot\left(1+2^4+...+2^{20}\right)\text{ }⋮\text{ }10\)

Vậy, \(C\text{ }⋮\text{ }10.\)

a) c = 2 + 2³ + 2⁵ + ... + 2¹⁹ + 2²¹ + 2²³

= (2 + 2³ + 2⁵) + (2⁷ + 2⁹ + 2¹¹) + ... + (2¹⁹ + 2²¹ + 2²³)

= 2.(1 + 2² + 2⁴) + 2⁷.(1 + 2² + 2⁴) + ... + 2¹⁹.(1 + 2² + 2⁴)

= 2.21 + 2⁷.21 + ... + 2¹⁹.21

= 21.(2 + 2⁷ + ... + 2¹⁹) ⋮ 21

Vậy c ⋮ 21

b) c = 2 + 2³ + 2⁵ + 2⁷ + ... + 2²¹ + 2²³

= (2 + 2³) + (2⁵ + 2⁷) + ... + (2²¹ + 2²³)

= 10 + 2⁴.(2 + 2³) + ... + 2²⁰.(2 + 2³)

= 10 + 2⁴.10 + ... + 2²⁰.10

= 10.(1 + 2⁴ + ... + 2²⁰) ⋮ 10

Vậy c ⋮ 10

a) \(A=2+2^2+2^3+...+2^{20}\)

\(A=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{19}+2^{20}\right)\)

\(A=2\cdot\left(1+3\right)+2^3\cdot\left(1+3\right)+...+2^{59}\cdot\left(1+3\right)\)

\(A=3\cdot\left(2+2^3+...+2^{59}\right)\)

Vậy A chia hết cho 3

________

\(A=2+2^2+2^3+...+2^{20}\)

\(A=\left(2+2^3\right)+\left(2^2+2^4\right)+...+\left(2^{58}+2^{60}\right)\)

\(A=2\cdot\left(1+4\right)+2^2\cdot\left(1+4\right)+...+2^{58}\cdot\left(1+4\right)\)

\(A=5\cdot\left(2+2^2+...+2^{58}\right)\)

Vậy A chia hết cho 5 

7)a) abcabc : abc = 1001 
abcabc = 1001 x abc . Mà 1001 chia hết cho 7; 11; 13 nên 1001 x abc chia hết cho 7; 11; 13 . Vậy abcabc chia hết cho 7; 11; 13 ( đpcm)
b .Vì abc = 2 . deg nên  abcdeg : deg = 2001
abcdeg = 2001 x deg. Do 2001 chia hết cho 23 và 29 nên 2001 x deg chia hết cho 23 và 29 . Vậy abcdeg chia hết cho 23 và 29 ( đpcm)
 

Ta có : 

abcabc = 1000abc + abc 

= 1001 . abc 

= 7 . 11 . 13 . abc chia hết cho 7 ; 11 ; 13

+, Ta có:

\(B=23!+19!-15!\)

\(B=\left(1\times2\times...\times11\times...\times23\right)+\left(1\times2\times...\times11\times...\times19\right)-\left(1\times2\times...\times11\times...\times15\right)\)

\(B=11\times\left[\left(1\times2\times...\times10\times12\times...\times23\right)+\left(1\times2\times...\times10\times12\times...\times19\right)-\left(1\times2\times...\times10\times12\times...\times15\right)\right]\)

\(\Rightarrow B⋮11\)

+, Ta có:

\(B=23!+19!-15!\)

\(B=\left(1\times2\times...\times10\times11\times...\times23\right)+\left(1\times2\times...\times10\times11\times...\times19\right)-\left(1\times2\times...\times10\times11\times...\times15\right)\)

\(B=11\times10\times\left[\left(1\times2\times...\times9\times12\times...\times23\right)+\left(1\times2\times...\times9\times12\times...\times19\right)-\left(1\times2\times...\times9\times12\times...\times15\right)\right]\)

\(B=110\times\left[\left(1\times2\times...\times9\times12\times...\times23\right)+\left(1\times2\times...\times9\times12\times...\times19\right)-\left(1\times2\times...\times9\times12\times...\times15\right)\right]\)

\(\Rightarrow B⋮110\)

+,Ta có:

\(B=23!+19!-15!\)

\(B=\left(1\times2\times...\times5\times...\times23\right)+\left(1\times2\times...\times5\times...\times19\right)-\left(1\times2\times...\times5\times...\times15\right)\)

\(B=5\times\left[\left(1\times2\times...\times4\times6\times...\times23\right)+\left(1\times2\times...\times4\times6\times...\times19\right)-\left(1\times2\times...\times4\times6\times...\times15\right)\right]\)

\(\Rightarrow B⋮5\)

~ Chúc bạn học tốt ~!

bbbbbbbbb

ccccccccc

Ta có: 7a+3b⋮23⇒6(7a+3b)⋮237a+3b⋮23⇒6(7a+3b)⋮23

⇒6(7a+3b)+(4a+5b)⋮23⇒6(7a+3b)+(4a+5b)⋮23

⇒46a+23b⋮23⇒23(2a+b)⋮23⇒46a+23b⋮23⇒23(2a+b)⋮23(Đúng)

Vậy 4a+5b⋮23