Question

Determinati cifrele nenule x,y,z ,stiind ca x

Answer 1

[tex]\overline{x,(yz)}+\overline{y,(zx)}+\overline{z,(xy)}=x+y+z+1\\ x\frac{\overline{yz}}{99}+y\frac{\overline{zx}}{99}+z\frac{\overline{xy}}{99}=x+y+z+1\\ \frac{99x+10y+z}{99}+\frac{99y+10z+x}{99}+\frac{99z+10x+y}{99}=x+y+z+1\\ \frac{99x+99y+99z+11x+11y+11z}{99}=x+y+z+1|\cdot99\\ 99x+99y+99z+11x+11y+11z=99x+99y+99z+99\\ 11x+11y+11z=99\\ 11(x+y+z)=99\\ x+y+z=9\\ Acum\ incerci\ toate\ variantele\ posibile\ tinand\ cond\ de:x\ \textless \ y\ \textless \ z[/tex]