Question

Determinati numerele naturale a,b,c,stiind ca 2 la puterea a+b+2 la putere b+c=40

Answer 1

$2^{a+b}$ ≤ 40 => a+b <6
a+b=0 => $2^{b+c} =39$, imposibil
a+b=1 => $2^{b+c} =38$, imposibil
a+b=2 => $2^{b+c} =36$, imposibil
a+b=3 => $2^{b+c} =32 => b+c=5$
a+b=4 => $2^{b+c} =24$, imposibil
a+b=5 => $2^{b+c} =8 => b+c=3$

Deci avem 2 cazuri:
I. a+b=3 si b+c=5, cu solutiile (a,b,c)∈{(0,3,2);(1,2,3);(2,1,4);(3,0,5)}
II. a+b=5 si b+c=3, cu solutiile (a,b,c)∈{(2,3,0);(3,2,1);(4,1,2);(5,0,3)}