Q35. 35. 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?

Let the two-digit number X be represented as 10a + b, where 'a' is the tens digit (1-9) and 'b' is the units digit (0-9). When digits are interchanged, Y = 10b + a. The sum X + Y = (10a + b) + (10b + a) = 11a + 11b = 11(a + b).

We are given that (X + Y) is the greatest two-digit number, which is 99. So, 11(a + b) = 99, which implies a + b = 9. We need to find the number of possible values for X, meaning the number of pairs (a, b) that satisfy a + b = 9, with 'a' from 1 to 9 and 'b' from 0 to 9.

Additionally, Y must also be a two-digit number, meaning 'b' cannot be 0. So, 'b' must be from 1 to 9. Possible pairs (a, b) where a ∈ [1,9] and b ∈ [1,9]: (1, 8) => X=18, Y=81 (2, 7) => X=27, Y=72 (3, 6) => X=36, Y=63 (4, 5) => X=45, Y=54 (5, 4) => X=54, Y=45 (6, 3) => X=63, Y=36 (7, 2) => X=72, Y=27 (8, 1) => X=81, Y=18 If a=9, then b=0, which would make Y=09=9, not a two-digit number.

So, (9,0) is not a valid pair. There are 8 possible values for X. This question tests number properties and careful interpretation of conditions.