Question

$(3^1 ^6 +3 ^{14} - 3 ^{13} )divizibil cu 29$

Answer 1

$(3^{16}+3^{14}-3^{13})~ \vdots ~29 \\ (3^1^6+3^1^4-3^1^3)=[3^{13}(3^3+3-1)]=[3^1^3(27+3-1)]= \\ =[3^1^3(30-1)]=(3^1^3 \cdot 29) ~\vdots ~29$