Suppose P and Q are distinct two-digit numbers consisting of the same

Suppose P and Q are distinct two-digit numbers consisting of the same digits. Then P-Q is

a prime number
an even number
an odd number
divisible by 9
This question was previously asked in
UPSC CISF-AC-EXE – 2019
The correct answer is (D) divisible by 9. Let the two distinct two-digit numbers be P and Q, formed by the digits ‘a’ and ‘b’. Since they are distinct two-digit numbers consisting of the same digits, the digits must be distinct (a ≠ b) and non-zero (otherwise one number would be a single digit). Let P = 10a + b and Q = 10b + a (where a and b are single digits from 1-9, a ≠ b).
– The difference P – Q = (10a + b) – (10b + a) = 9a – 9b = 9(a – b).
– Since ‘a’ and ‘b’ are distinct digits, (a – b) is a non-zero integer.
– Therefore, the difference P – Q is always a multiple of 9.
– Any multiple of 9 is divisible by 9.
Let’s test the other options with an example: P=72, Q=27. P-Q = 72-27 = 45.
– Is 45 a prime number? No.
– Is 45 an even number? No.
– Is 45 an odd number? Yes. But consider P=31, Q=13. P-Q = 31-13 = 18, which is even. So it’s not always odd.
– Is 45 divisible by 9? Yes (45 / 9 = 5).
The property P-Q = 9(a-b) guarantees divisibility by 9 for any pair of distinct two-digit numbers formed by swapping two distinct digits (assuming both digits are non-zero to ensure two distinct two-digit numbers, or carefully considering the case with digit 0, which would mean numbers like 20 and 02, but 02 is not a two-digit number, confirming our assumption that both digits must be non-zero).