Question

Se considera x,y,z numere naturale astfel incat 9x-5=13y+4z. Determinati restul impartirii sumei x+z la 13 si restul impartirii sumei y+z la 9.

Answer 1

Explicație pas cu pas:

$9x - 5 = 13y + 4z$

$9x + 9z = 13y + 4z + 9z + 5 \\ 9(x + z) = 13(y + z) + 5$

$x + z = \dfrac{13(y + z) + 5}{9}$

$\dfrac{x + z}{13} = \dfrac{13(y + z) + 5}{9 \cdot 13} = \dfrac{y + z}{9} + \dfrac{5}{9 \cdot 13}$

$9(x + z) - 5 = 13(y + z)$

$y + z = \dfrac{9(x + z) - 5}{13}$

$\dfrac{y + z}{9} = \dfrac{9(x + z) - 5}{13 \cdot 9} = \dfrac{x + z}{13} - \dfrac{5}{13 \cdot 9}$