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

Răspuns :

[tex] \frac{2x+1}{13} [/tex] ∈ 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 ; .. }
[tex]\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\\ \\[/tex]

[tex] $ \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}}[/tex]


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