Question

Determinați nr nat a și b, astfel încât (a+b)•(b-3)=28

Answer 1

$(a+b)\cdot (b-3) = 28 ,\quad a,b\in \mathbb_{N} \\ \\ \bullet (a+b) \in D_{28}^+ \Rightarrow (a+b) \in\Big\{1,2,7,14,28\Big\} \\ \\ \bullet (b-3)\in D_{28}^+ \Rightarrow (b-3)\in\Big\{1,2,7,14,28\Big\} \\ \\ Dar, (a+b)\cdot (b-3) = 28 \Rightarrow \\ \\ \Rightarrow \boxed{1} \quad (a+b) \in \Big\{1,2,7,14,28\Big\} si (b-3)\in \Big\{28,14,7,2,1\Big\}\Big|+3 \\ (in ordinea asta) \\ \\ \Rightarrow (a+b)\in \Big\{1,2,7,14,28\Big\} si b \in \Big\{31,17,10,5,4\Big\}$

$(a+b) \in \Big\{{1,2,7,14,28\Big\} \Big|-b \Rightarrow \\ \\ \Rightarrow a\in \Big\{1-31,2-17,7-10,14-5,28-4\Big\} \Rightarrow \\ \\ \Rightarrow a \in \Big\{-30,-15,-3,9,24\Big\}, dar, a\in \mathbb_{N} \Rightarrow a\in \Big\{9,24\Big\} \Rightarrow \\ \\ \Rightarrow ( luam valorile lui b de pe pozitiile 4 si 5) \Rightarrow b \in \Big\{5,4\Big\} \\ \\ \Rightarrow (a,b) = \Big\{(9,5); (24,4)\Big\}$

$\boxed{2} \quad (a+b) \in\Big\{28,14,7,2,1\Big\} si (b-3)\in \Big\{1,2,7,14,28\Big\}\Big|+3 \\ \\ \Rightarrow \quad b \in \Big\{4,5,10,17,31\Big\} \\ \\ (a+b) \in \Big\{28,14,7,2,1\Big\} \Big|-b \Rightarrow a\in \Big\{24,9,-3,-15-30\Big\}, dar a\in \mathbb_N \Rightarrow \\ \\ \Rightarrow a\in \Big\{24,9\Big\} \Rightarrow ( luam termenii de pe pozitiile 1 si 2 din b) \Rightarrow \\ \\ \Rightarrow b \in \Big\{4,5\Big\} \Rightarrow (a,b) = \Big\{(24,4);(9,5)\Big\}$

$\\ Din \boxed{1} \cup \boxed{2} \Rightarrow (a,b)\in \Big\{(9,5);(24,4);(24,4);(9,5)\Big\} \Rightarrow \\ \\ \Rightarrow \boxed{(a,b) \in \Big\{(9,5);(24,4)\Big\}}$