arătați ca 11|A, unde A=
[tex] {8}^{n + 1} \times {5}^{3n} + {2}^{3n + 2} \times 125 - {4}^{n} \times {5}^{2n} \times {10}^{n} [/tex]
, pentru orice n aparține numerelor naturale nenule.
VA ROG E URGENT DAU 50 DE PUNCTE+COROANĂ​

Răspuns :

Răspuns:

[tex] \large \bf A={8}^{n + 1} \cdot {5}^{3n} +{2}^{3n+2} \cdot \: 125 - {4}^{n} \cdot {5}^{2n} \cdot {10}^{n}[/tex]

[tex]\large \bf A={2}^{3n+3} \cdot {5}^{3n} +{2}^{3n+2} \cdot {5}^{3}-{2}^{2n} \cdot {5}^{2n} \cdot {10}^{n}[/tex]

[tex]\large \bf A={10}^{3n} \cdot {2}^{3} +{10}^{3n}\cdot {2}^{2}-{10}^{2n}\cdot {10}^{n}[/tex]

[tex]\large \bf A={10}^{3n} \cdot {2}^{3} +{10}^{3n}\cdot {2}^{2}-{10}^{3n}[/tex]

[tex]\large \bf A={10}^{3n} \cdot ({2}^{3} + {2}^{2}-{10}^{0})[/tex]

[tex]\large \bf A={10}^{3n} \cdot (8+ 4-1)[/tex]

[tex]\large \bf A={10}^{3n} \cdot 11 \implies \boxed{ \bf\:A \: \vdots \: 11}[/tex]

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