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?

The correct answer is None of the above.

Step-by-step Explanation:

Let a 2-digit number be represented as $10x+y$, where $x$ is the tens digit and $y$ is the units digit. Note that since it is a 2-digit number, $x$ must be an integer from 1 to 9 ($1\le x\le 9$), and $y$ must be an integer from 0 to 9 ($0\le \mathrm{y<mo>\le </mo><mn>9</mn>}$).

When the digits are reversed, the new number becomes $10y+x$.

According to the question, the difference between the original number and the reversed number is always 27. So, we can set up the equation:

$(10x+y)-(10y+x)=27$

Simplifying the left side:

$9x-9y=27$

Dividing the entire equation by 9 gives:

$x-y=3$

Or equivalently, $x=y+3$.

Now, let us find all valid pairs of digits $(x,y)$ such that $1\le x\le 9$ and $0\le \mathrm{y<mo>\le </mo><mn>9</mn>}$:

1. If $y=0$, then $x=3$ → Number is 30 (Difference: 30 - 03 = 27)
2. If $y=1$, then $x=4$ → Number is 41 (Difference: 41 - 14 = 27)
3. If $y=2$, then $x=<mn>5</mn>$ → Number is 52 (Difference: 52 - 25 = 27)
4. If $y=3$, then $x=<!--</mo--><mn>6</mn>$ → Number is 63 (Difference: 63 - 36 = 27)
5. If $y=4$, then $x=7$ → Number is 74 (Difference: 74 - 47 = 27)
6. If $y=5$, then $x=8$ → Number is 85 (Difference: 85 - 58 = 27)
7. If $y=6$, then $x=9$ → Number is 96 (Difference: 96 - 69 = 27)

Thus, there are a total of 7 such 2-digit numbers: 30, 41, 52, 63, 74, 85, and 96.

Since 7 is not listed among options 3, 4, or 5, the correct choice is None of the above.