Question

Determinati x,a,b. numere naturale daca 2⁵ˣ⁺¹+2²ᵃ+2ᵇ=2368.

Answer 1

Explicație pas cu pas:

2368 = 2⁶×37

${2}^{5x + 1} + {2}^{2a} + {2}^{b} = {2}^{6} \cdot 37$

${2}^{6} \cdot ({2}^{5x - 5} + {2}^{2a - 6} + {2}^{b - 6}) = {2}^{6} \cdot 37$

${2}^{5(x - 1)} + {2}^{2(a - 3)} + {2}^{b - 6} = 37$

unul din termeni este impar => 2⁰ = 1

atunci:

${2}^{6} \cdot (1 + 4 + 32) = {2}^{6} \cdot 37\\ {2}^{6} \cdot ({2}^{0} + {2}^{2} + {2}^{5}) = {2}^{6} \cdot 37 \\ \iff {2}^{6} + {2}^{8} + {2}^{11} = 2368$

dacă:

$5x + 1 = 6 \iff 5x = 5 \\ \implies x = 1$

2a nu poate fi impar =>

$2a = 8 \implies a = 4$

$\implies b = 11$

dacă:

$5x + 1 = 8 \iff 5x = 7 \\ \implies x \not \in \mathbb{N}$

dacă:

$5x + 1 = 11 \iff 5x = 10 \\ \implies x = 2$

$2a = 6 \implies a = 3 \implies b = 8$

sau

$2a = 8 \implies a = 4 \implies b = 6$

=> numerele sunt:

(x,a,b) ∈ {(1,4,11); (2,3,8); (2,4,6)}