# Q. 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?

a) 2

b) 4

c) 6

d) 8

Correct answer: d) 8

#### Question from UPSC Prelims 2024 CSAT

**Explanation : **

## Two-Digit Number Sum by Interchanging Digits of X

Let’s analyze the problem step by step.

**Given:**

– Let X = 10a + b be a two-digit number where a is the tens digit (1-9) and b is the units digit (0-9).

– Let Y = 10b + a be the number formed by interchanging the digits of X.

**Sum of X and Y:**

X + Y = (10a + b) + (10b + a) = 11(a + b)

**Objective:**

– To maximize X + Y such that it remains a two-digit number.

Maximum Two-Digit Number:

– The greatest two-digit number is 99.

– So, 11(a + b) ≤ 99 which implies a + b ≤ 9.

**Maximizing the Sum:**

– To achieve the largest possible two-digit sum, set a + b = 9.

– Thus, X + Y = 99.

**Possible Combinations:**

We need to find pairs (a, b) where a + b = 9 and a ranges from 1 to 9 while b ranges from 0 to 9.

**Here are the possible pairs:**

(1,8), (2,7), (3,6), (4,5), (5,4), (6,3), (7,2), (8,1), (9,0)

**Excluding Invalid Cases:**

– When a = 9, b = 0 leads to Y = 09, which is not a valid two-digit number.

– Therefore, exclude (9,0).

**Valid Pairs:**

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

Number of Possible Values of X:

– There are 8 valid pairs.

Answer: d) 8