OSN 2011 Problem 1

For a number $n$ in base $10$ , let $f(n)$ be the sum of all numbers possible by removing some digits of $n$ (including none and all). For example, if $n = 1234$ , $f(n) = 1234 + 123 + 124 + 134 + 234 + 12 + 13 + 14 + 23 + 24 + 34 + 1 + 2 + 3 + 4 = 1979$ ; this is formed by taking the sums of all numbers obtained when removing no digit from $n$ (1234), removing one digit from $n$ (123, 124, 134, 234), removing two digits from $n$ (12, 13, 14, 23, 24, 34), removing three digits from $n$ (1, 2, 3, 4), and removing all digits from $n$ (0). If $p$ is a 2011-digit integer, prove that $f(p)-p$ is divisible by $9$ .

Remark: If a number appears twice or more, it is counted as many times as it appears. For example, with the number $101$ , $1$ appears three times (by removing the first digit, giving $01$ which is equal to $1$ , removing the first two digits, or removing the last two digits), so it is counted three times.