E.127. Determinați cardinalul mulțimii A={ab‾ ∣ 16a2+b∈N}A=\bigg\{ \overline{ab} ~ \bigg | ~ \dfrac{16}{a^2+b} \in \N \bigg\}A={ab a2+b16∈N}.
Indicații: 16a2+b∈N⇒a2+b∈{1,2,4,8,16}\dfrac{16}{a^2+b} \in \N \Rightarrow a^2+b \in \{1, 2, 4, 8, 16\}a2+b16∈N⇒a2+b∈{1,2,4,8,16}.
Răspuns: card(A)=8card(A)=8card(A)=8.
16a2+b∈N⇒a2+b∈{1,2,4,8,16}\dfrac{16}{a^2+b} \in \N \Rightarrow a^2+b \in \{1, 2, 4, 8, 16\}a2+b16∈N⇒a2+b∈{1,2,4,8,16}.
∙a2+b=1⇒ab‾∈{10}\bullet \quad a^2+b = 1 \Rightarrow \overline{ab} \in \{10 \}∙a2+b=1⇒ab∈{10};
∙a2+b=2⇒ab‾∈{11}\bullet \quad a^2+b = 2 \Rightarrow \overline{ab} \in \{11 \}∙a2+b=2⇒ab∈{11};
∙a2+b=4⇒ab‾∈{13,20}\bullet \quad a^2+b = 4 \Rightarrow \overline{ab} \in \{13, 20 \}∙a2+b=4⇒ab∈{13,20};
∙a2+b=8⇒ab‾∈{17,24}\bullet \quad a^2+b = 8 \Rightarrow \overline{ab} \in \{17, 24 \}∙a2+b=8⇒ab∈{17,24};
∙a2+b=16⇒ab‾∈{37,40}\bullet \quad a^2+b = 16 \Rightarrow \overline{ab} \in \{37, 40 \}∙a2+b=16⇒ab∈{37,40};
Deci A={10,11,13,20,17,24,37,40}⇒card A=8A=\{10, 11, 13, 20, 17, 24, 37, 40\} \Rightarrow \boxed{card~A=8}A={10,11,13,20,17,24,37,40}⇒card A=8.