elementary number theory - Proving that $9$ is a divisor of $x in Bbb N$ if the sum of digits of $x$ is divisible by $9$.

Suppose x is a positive integer with $n$ digits, say $x = d_1d_2d_3\ldots d_n.$ If $9$ is a divisor of $d_1 + d_2 + \ldots d_n$, prove then $9$ is a divisor of $x$.

My attempt: suppose $x = 4518.$ Therefore $d_1 = 4, d_2 = 5, d_3 = 1, d_4 = 8$ those added together equals $18$, where $9$ is a divisor.

With that in mind $4518$ can be written as $4000 + 500 + 10 + 8.$ How do you show that $9$ is a divisor of this entire number from the information given?