### Instruction: What is the smallest 9 digit number of the form $aaa, \! aaa, \! aaa$ such that it is divisible by 9? (Notice that all the digits of this number must be the same!).### Response: To solve this problem, I need to recall the divisibility rule for 9.
A number is divisible by 9 if and only if the sum of its digits is divisible by 9.
So, I need to find the smallest digit a such that $9a$ is divisible by 9.
Clearly, the smallest such digit is 1, because $9 \times 1 = 9$, which is divisible by 9.
Therefore, the smallest 9 digit number of the form $aaa, \! aaa, \! aaa$ that is divisible by 9 is $111,111,111$.