Câu hỏi của mr. killer

Question

Cho 10k-1⋮19(k∈N)CMR:

a) $10^{2k}-1\vdots 19$

b) $10^{3k}-1\vdots 19$

Lightning Farron · Accepted Answer

a)$10^{2k}-1=\left(10^k-1\right)\left(10^k+1\right)$

Dễ thấy: $10^k-1⋮19\Rightarrow\left(10^k-1\right)\left(10^k+1\right)⋮19$

$\Rightarrow10^{2k}-1⋮19$

b)$10^{3k}-1=\left(10^k-1\right)\left(10^k+10^{2k}+1\right)$

Dễ thấy: $10^k-1⋮19\Rightarrow\left(10^k-1\right)\left(10^k+10^{2k}+1\right)⋮19$

$\Rightarrow10^{3k}-1⋮19$Câu trả lời: a)$10^{2k}-1=\left(10^k-1\right)\left(10^k+1\right)$

Dễ thấy: $10^k-1⋮19\Rightarrow\left(10^k-1\right)\left(10^k+1\right)⋮19$

$\Rightarrow10^{2k}-1⋮19$

b)$10^{3k}-1=\left(10^k-1\right)\left(10^k+10^{2k}+1\right)$

Dễ thấy: $10^k-1⋮19\Rightarrow\left(10^k-1\right)\left(10^k+10^{2k}+1\right)⋮19$

$\Rightarrow10^{3k}-1⋮19$

Eren · Answer

Thắng xem mà học tập đây :v

Vì 10k - 1 $⋮$ 19 => 10k - 1$\equiv$ 0 (mod 19)

=> 10k $\equiv$ 1 (mod 19)

a) 10k $\equiv$ 1 (mod 19)

=> (10k)2 $\equiv$ 12 (mod 19)

=> 102k $\equiv$ 1 (mod 19)

=> 102k - 1 $⋮$ 19

b) 10k $\equiv$ 1 (mod 19)

=> (10k)3 $\equiv$ 13 (mod 19)

=> 103k = 1 (mod 19)

=> 103k - 1 $⋮$ 19Câu trả lời: Thắng xem mà học tập đây :v