Question

Se considera numarul x=a,b(c)+b,c(a)+c,a(b) supra a+b+c , unde abc apartin {1,2,3,4,5,6,7,8} si a diferit de b diferit de c diferit de a. Aratati ca 9x este un numar natural. URGENT!! DAU COROANA SI 52 DE PUNCTE

Answer 1

[tex]\it x = \dfrac{\overline{a,b(c)} + \overline{b,c(a)} + \overline{a,b(c)}}{a+b+c} \Rightarrow 10x = \dfrac{\overline{ab,(c)} + \overline{bc,(a)} + \overline{ab,(c)}} {a+b+c} = \\ \\ \\ = \dfrac{\overline{ab} +0,(c) +\overline{bc} +0,(a) +\overline{ca} +0,(b)}{a+b+c}=\\ \\ \\ = \dfrac{10a+b +\dfrac{c}{9} +10b+c +\dfrac{a}{9} +10c+a +\dfrac{b}{9}}{a+b+c}=[/tex]

[tex]\it =\dfrac{ 11a+11b+11c+\dfrac{a+b+c}{9}}{a+b+c} = \dfrac{ 11(a+b+c)+\dfrac{a+b+c}{9}}{a+b+c}=\\ \\ \\=\dfrac{(a+b+c)(11+\dfrac{1}{9})}{a+b+c} = \dfrac{100}{9} \\ \\ \\ 10x= \dfrac{100}{9} \Rightarrow x = \dfrac{10}{9} \\ \\ \\ 9x= 9\cdot\dfrac{10}{9} =10 \in \mathbb{N}[/tex]

Answer 2

Am atasat rezolvarea.