Question

if xyz+xy+yz+zx+x+y+z=84where x,y,zare positive integeres then the value of (x+y+z)is

Answer 1

xyz+xy+xz+yz+x+y+z=384xy (z+1) + z (x+y)+1(x+y)+z=384xy (z+1) + (x+y) (z+1) +(z+1)−1=384xy (z+1) + (x+y) (z+1) +(z+1)=385(z+1) (xy+x+y+1)=385(z+1) (x(y+1)+1(y+1))=385(x+1)(y+1)(z+1)=385(x+1)(y+1)(z+1)=5×7×11On comparing ,x+1 = 5 , y+1=7 , z+1=11or x+1 = 7 , y+1=11 , z+1=5or x+1 = 11 , y+1=5 , z+1=7Now , For each casex+1+y+1+z+1=5+11+7x+y+z+3=23x+y+z=20