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?

Correct Option (D)

Let the two-digit number be represented as 10x + y, where x is the tens digit and y is the units digit. The number obtained by reversing its digits is 10y + x.

According to the problem statement, the difference between the original number and the reversed number is 27:

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

Simplifying the equation:

10x + y - 10y - x = 27

9x - 9y = 27

Dividing both sides by 9:

x - y = 3

For a two-digit number, x must be a digit from 1 to 9 (as the tens digit cannot be 0), and y must be a digit from 0 to 9. We need to find pairs of (x, y) that satisfy the condition x - y = 3:

• If y = 0, then x = 3. The number is 30. (30 - 03 = 27)
• If y = 1, then x = 4. The number is 41. (41 - 14 = 27)
• If y = 2, then x = 5. The number is 52. (52 - 25 = 27)
• If y = 3, then x = 6. The number is 63. (63 - 36 = 27)
• If y = 4, then x = 7. The number is 74. (74 - 47 = 27)
• If y = 5, then x = 8. The number is 85. (85 - 58 = 27)
• If y = 6, then x = 9. The number is 96. (96 - 69 = 27)

There are 7 such two-digit numbers that satisfy the given condition. Since 7 is not listed in options 1, 2, or 3, option 4, "None of the above," is the correct answer.

Incorrect Options:

Options 1, 2, and 3 propose that there are 3, 4, or 5 such two-digit numbers, respectively. This is incorrect because, as demonstrated by the derivation x - y = 3 and the subsequent enumeration of all valid digit pairs, there are precisely 7 numbers that meet the specified criteria. Therefore, these options do not represent the correct maximum count of such numbers.