Let the two-digit number X be represented as $10a + b$, where a is the tens digit and b is the ones digit.

The number Y, formed by interchanging the digits of X, will be $10b + a$.

The sum X + Y = $(10a + b) + (10b + a) = 11a + 11b = 11(a + b)$.

We are told that X + Y is the greatest two-digit number, which is 99. So, $11(a + b) = 99$.

This simplifies to $a + b = 9$.

Now, a and b are digits (ranging from 0 to 9) and must satisfy $a + b = 9$. The possible pairs for (a, b) are:

• (1, 8)
• (2, 7)
• (3, 6)
• (4, 5)
• (5, 4)
• (6, 3)
• (7, 2)
• (8, 1)

Thus, there are 8 possible values for X.

Answer: D. 8