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: \(10^k-1⋮19\Rightarrow\left(10^k-1\right)\left(10^k+10^{2k}+1\right)⋮19\)
\(\Rightarrow10^{3k}-1⋮19\)
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 \(⋮\) 19
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
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
