X=a+(a+ 1)+...+(a+8)= 9a+36 = 9(a+4)；

X=b+(b+ 1)+...+(b+9)= 10 b+45 = 5(2 b+9)；

X=c+(c+ 1)+...+(c+ 10)= 1 1c+55 = 1 1(c+5)；

So, as long as x is a multiple of 9, 5, 1 1.