Răspuns:
Explicație pas cu pas:
[tex]notez\:cu\:q_{1},\:respectiv\:q_{2}\:cele\:dou\breve{a}\:c\hat{a}turi.[/tex]
[tex]121\div n=q_{1}\:rest\:13\implies121-13=n\cdot q_{1}\implies108=n\cdot q_{1}\implies n\:|\:108\\283\div n=q_{2}\:rest\:13\implies283-13=n\cdot q_{2}\implies270=n\cdot q_{2}\implies n\:|\:270[/tex]
Calculăm (108;270), unde am notat cu (a;b) cel mai mare divizor comun al numerelor a și b.
[tex]108={2}^{2}\cdot{3}^{3}\\270=2\cdot{3}^{3}\cdot5\\\implies(108;270)=2\cdot{3}^{3}-2\cdot27=\bold{54}[/tex]
[tex]\implies n\in D_{54}\implies \bold{n\in\left\{1;2;3;6;9;18;27;54\right\}}[/tex]
