UPSC Prelims 2026 CSAT, question 66

Topic: Number system

Q66

$A$ is a 2-digit number with different digits. $B$ is also a 2-digit number and is obtained by reversing the digits of $A$ . If $A-B$ is a multiple of 27, where $A>B$ , how many such different $A$ 's are possible?

(a) 6
(b) 9Correct answer
(c) 12
(d) 18

Solution.

Answer: (b) 9

Let $A = 10x + y$ and $B = 10y + x$ with distinct digits $x, y$ ; since $B$ is a 2-digit number and $A > B$ , we have $1 \le y < x \le 9$ .
Then $A - B = 9(x - y)$ is a multiple of $27$ if and only if $x - y$ is a multiple of $3$ , so $x - y \in \{3, 6\}$ .
For $x - y = 3$ , $y \in \{1, 2, 3, 4, 5, 6\}$ gives $6$ numbers; for $x - y = 6$ , $y \in \{1, 2, 3\}$ gives $3$ numbers, making $6 + 3 = 9$ possible values of $A$ .

Trap: Because $B$ is explicitly stated to be a 2-digit number, the units digit $y$ of $A$ cannot be $0$ (which would add $30, 60, 90$ to give 12).