Question

Care este multimea continand x apartinand N, cu proprietatea 2x+1/13apartine N?

Answer 1

$\frac{2x+1}{13}$ ∈ n ⇒ 2x+1 ∈M₁₃⇒
⇒2x+1 ∈ {13 ; 26 ; 39 ; 52 ; ... }  | -1
2x ∈ {12 ; 25 ; 38 ; 51 ; .. }   | :2
x ∈ { 6; 25/2 ; 19 ; 51/2 ; ... } => x ={ 6 ; 19 ; 32 ; .. }

Answer 2

$\dfrac{2x+1}{13}\in \mathbb_{N} \Rightarrow 2x+1 \in M_{13} \Rightarrow 2x+1\in \Big\{13,26,39,52,...\Big\} \Rightarrow \\ \\ \Rightarrow 2x+1 = 13\cdot n,\quad n\in \mathbb_{N} *\\ \\ 2x+1 = 13\cdot n \Big|-1\Rightarrow 2x = 13\cdot n-1 \Rightarrow x = \dfrac{13\cdot n-1}{2} \\ \\ x\in \mathbb_{N} \Rightarrow 13\cdot n-1 par \Rightarrow 13\cdot n -1 par doar atunci cand \\ 13\cdot n impar \Rightarrow 13\cdot n impar doar atunci cand n impar$

$\Rightarrow n \in \Big\{1,3,5,7,9,11,...\Big\}\\ \\ \Rightarrow \x = \dfrac{13\cdot n-1}{2},\quad n\in \Big\{1,3,5,7,9,11,...\Big\} \Rightarrow \\ \\ \Rightarrow x\in \Big\{\dfrac{13-1}{2},\dfrac{39-1}{2},\dfrac{65-1}{2},\dfrac{91-1}{2},\dfrac{117-1}{2},\dfrac{143-1}{2},...\Big\} \Rightarrow \\ \\ \Rightarrow \boxed{x\in \Big\{6,19,32,45,58,71...\Big\}} \Rightarrow \\ \\ \\ \Rightarrow \boxed{x= \dfrac{13\cdot n-1}{2}, \quad n \in \mathbb_{N} \backslash 2\mathbb_{N}}$


$\Big(\mathbb_{N} \backslash 2\mathbb_{N} \text{este multimea numerelor naturale impare.}\Big)$