Question

Ex 24 punctele b si c . Va rog!!​

Answer 1

$\displaystyle\bf\\b)~B=(6^{80}+2^{81}\cdot3^{79}) \vdots 5,~folosim~criteriul~de~divizibilitate~cu~5~:\\un~numar~este~divizibil~cu~5~daca~ultima~sa~cifra~este~0~sau~5.\\deci~in~cazul~asta~trebuie~sa~aflam~U(6^{80} +2^{81}\cdot 3^{79})=\\U(6^{80})+U(2^{81})\cdot U(3^{79}) = \overline{...6} + \overline{...2}\cdot \overline{...7}=\overline{...0} \implies B \vdots 5.$

$\displaystyle\bf\\c)~C=(4^{15}+3\cdot 2^{28})\vdots 7,~aici~vom~incerca~sa~rezolvam~altfel,~fara~criterii.\\C=2^{30}+3\cdot 2^{28}=2^{28}(2^2+3)=2^{28} \cdot 7 \Leftrightarrow C \vdots 7.$