Question

Cho 10^k-1⋮19(k∈N)CMR:

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

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

Answer 1

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

\(\Rightarrow10^{3k}-1⋮19\)

Answer 2

Thắng xem mà học tập đây :v

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

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

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

=> (10^k)² \(\equiv\) 1² (mod 19)

=> 10^2k \(\equiv\) 1 (mod 19)

=> 10^2k - 1 \(⋮\) 19

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

=> (10^k)³ \(\equiv\) 1³ (mod 19)

=> 10^3k = 1 (mod 19)

=> 10^3k - 1 \(⋮\) 19

Answer 3

Vì \(10^k-1⋮9\)

\(\Rightarrow10^k-1=19n\left(n\in N\right)\)

\(\Rightarrow10^k=19n+1\)

\(\Rightarrow10^k.10^k=\left(19n+1\right)\left(19n+1\right)\)

\(\Rightarrow10^{2k}=19^2n^2+2.19n+1\)

\(\Rightarrow10^{2k}-1=19^2n^2+2.19n⋮19\)

Câu còn lại tương tự

Answer 4

Vì \(10^k-1⋮19\)

\(\Rightarrow10^k-1=19n\left(n\in N\right)\)

\(\Rightarrow10^k=19n+1\)

\(\Rightarrow10^k.10^k=\left(19n+1\right)\left(19n+1\right)\)

\(\Rightarrow10^{2k}=19^2n^2+2.19n+1\)

\(\Rightarrow10^{2k}-1=19^2n^2+2.19n⋮19\)

Câu còn lại tương tự