Let X be a two-digit number and Y be another two-digit number formed by interchanging the digits of X. If (X + Y) is the greatest two-digit number, then what is the number of possible values of X?

The correct option is 8.

Let the two-digit number $X$ be represented as $10a+b$, where $a$ is the tens digit and $b$ is the units digit. Since $X$ is a two-digit number, the tens digit $a$ must be a non-zero single digit (i.e., $a\in \{1,2,3,4,5,6,7,8,9\}$), and the units digit $b$ is a single digit (i.e., $b\in \{0,1,2,3,4,5,6,7,8,9\}$).

The number $Y$ is formed by interchanging the digits of $X$, so we have:
$Y=10b+a$
Since $Y$ is also specified to be a two-digit number, its tens digit $b$ must also be non-zero (i.e., $b\ne 0$). Thus, both $a$ and $b$ can only take integer values from 1 to 9.

Now, let us calculate the sum $X+Y$:
$X+Y=(10a+b)+(10b+a)$
$X+Y=11(a+b)$

We are given that $X+Y$ is the greatest two-digit number. The greatest two-digit number is 99. Therefore:
$11(a+b)=99$
Dividing both sides by 11 gives:
$a+b=9$

Since $a$ and $b$ are positive single-digit integers (from 1 to 9), we can find all possible pairs $(a,b)$ that satisfy this equation:
1) $a=1,b=8\Rightarrow X=18$
2) $a=2,b=7\Rightarrow X=27$
3) $a=3,b=6\Rightarrow X=36$
4) $a=4,b=5\Rightarrow X=45$
5) $a=5,b=4\Rightarrow X=54$
6) $a=6,b=3\Rightarrow X=63$
7) $a=7,b=2\Rightarrow X=72$
8) $a=8,b=1\Rightarrow X=81$

Note that if $a=9$, then $b$ would have to be 0, which would make $Y=09$ (not a two-digit number). Thus, there are exactly 8 valid combinations. Therefore, the number of possible values of $X$ is 8.