Determine the three smallest positive integers that contain only \(0\)s and \(1\)s as digits and are divisible by \(6\).
A number divisible by \(6\) must be divisible by both \(2\) and \(3\).
Since the number is divisible by \(2\), it must be even. Since the only digits in our number are \(1\)s and \(0\)s, the units digit must be \(0\).
If a number is divisible by \(3\), the sum of the digits must also be divisible by \(3\). Since the only digits in our number are \(1\)s and \(0\)s, the smallest positive integer divisible by \(3\) must contain three \(1\)s.
To be as small as possible, the number must contain as few digits as possible. Therefore, the smallest such number contains three \(1\)s and ends in a \(0\). This number is \(1110\).
There are no more four-digit positive integers that end in a \(0\) and contain three \(1\)s. Therefore, the next smallest number will have five digits. To be divisible by \(3\) and contain five digits, which are only \(1\)s and \(0\)s, the number must contain three \(1\)s and two \(0\)s. As stated above, one \(0\) must be the units digit. To be as small as possible, the second \(0\) must be in the highest place value possible. Therefore, the next two smallest positive integers are \(10110\), and \(11010\).
Thus, the three smallest positive integers to meet the requirements are \(1110\), \(10110\), and \(11010\).