Question 1023878
Let x,y,and z be the digits:
100x+10y+z=17(x+y+z)
100x+10y+z+198=100z+10y+x
x+z+1=y;
So:
100x+z+198=100z+x
99x+198=99z
x+2=z
x+(x+2)+1=y
y=2x+3
100(x)+10(2x+3)+(x+2)=17(x+2x+3+x+2)
100x+20x+30+x+2=68x+85
53x+32=85
53x=53
x=1,y=5,z=3!!!!!!!!!!!!!!