Restul împărțirii numerului 4la puterea n x 18 la puterea n + 3 x(3²n-¹ +1) - 2³n la 9 la puterea n - 1 este a) 5, b) 4, c) 2, d) 1​

[tex]4^n\times18^n+3(3^{2n-1}+1)-2^{3n}:9^n-1[/tex]

• Calculam separat partea din stanga:

[tex]2^{2n}\times (3^2\times 2)^n+3(3^{2n-1}+1)-2^{3n}=[/tex]

[tex]2^{3n}\times 3^{2n}+3^{2n}+3-2^{3n}=2^{3n}(3^{2n}-1)+3^{2n}-1+4=[/tex]

[tex](3^{2n}-1)(2^{3n}+1)+4[/tex]

• Deci, vom avea:

[tex](3^{2n}-1)(2^{3n}+1)+4:9^n-1\\\\(3^{2n}-1)(2^{3n}+1)+4:3^{2n}-1[/tex]

Observam ca primul termen (3²ⁿ-1)(2³ⁿ+1) se imparte exact la 3²ⁿ-1, deci restul=4

[tex]\frac{(3^{2n}-1)(2^{3n}+1)+4}{3^{2n}-1}=\frac{(3^{2n}-1)(2^{3n}+1)}{3^{2n}-1} +\frac{4}{3^{2n}-1} =2^{3n}+1+\frac{4}{3^{2n}-1}[/tex]

Restul=4