Question

Va rog 13 si 14 dau coroana ​

Answer 1

Răspuns:

13. $x\in\{11, 24, 39\}$

Explicație pas cu pas:

13.

$x : 10 = c \: rest \: c^2, c^2 < 10, c\in \mathbb{N} \iff c \leq 3\\\\x = 10c + c^2 = c(c+10)\\\\c = 0 \implies x = 0(0+10) = 0 \textrm{ nu il punem la multimea solutiilor}\\\\c=1\implies x = 1(1+10) = 11\\\\c = 2 \implies x = 2(2+10) = 2\cdot 12 = 24\\\\c = 3\implies x = 3(3+10) = 3\cdot 13 = 39\\\\x\in\{11, 24, 39\}$

14.

$a=2^{n+2}\cdot 3^{n+2} + 5\cdot 6^n + 2^{n+1} \cdot 3^{n+1}\\\\a = (2\cdot 3)^{n+2} + 5\cdot 6^n + (2\cdot 3)^{n+1}\\\\a = 6^{n+2} + 5\cdot 6^n + 6^{n+1}\\\\a = 6^n(6^2 + 5 + 6)\\\\a = 6^n(36+11)\\\\a=6^n\cdot 47\\\\\implies a \in M_{47}\implies 47 \mid a, \forall n \in \mathbb{N}$

Answer 2

Explicație pas cu pas:

a=2^(n+2)*3^(n+2)+5*6^n+2^(n+1)*3^(n+1)

a=2^n*4*3^n*9+5*6^n+2^n*2*3^n*3

a=6^n*36+5*6^n+6^n*6

a=6^n(36+5+6)

a=6^n*47 ⇒a divizibil cu 47 oricare ar fi n∈N

Bafta!