There are certain 2-digit numbers. The difference between the number and the one obtained on reversing it is always 27. How many such maximum 2-digit numbers are there?

• Let the 2-digit number be represented as $10a + b$, where $a$ is the tens digit and $b$ is the units digit.

• The number obtained by reversing the digits is $10b + a$.

• The problem states the difference between the number and its reverse is 27, so $|(10a + b) - (10b + a)| = 27$.

• Simplifying the expression gives $|9a - 9b| = 27$, which reduces to $|a - b| = 3$.

• We must find pairs of digits $(a, b)$ where $a \in \{1, ..., 9\}$, $b \in \{0, ..., 9\}$, and the absolute difference is 3.

• The valid pairs for $(a, b)$ are: $(1,4), (2,5), (3,6), (4,7), (5,8), (6,9)$ where $a > b$, and $(4,1), (5,2), (6,3), (7,4), (8,5), (9,6)$ where $b > a$.

• This yields 12 distinct 2-digit numbers: 14, 25, 36, 47, 58, 69, 41, 52, 63, 74, 85, 96.

• Since 12 is not equal to 3, 4, or 5, the correct option is 'None of the above'.