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Q4546/80Q47
Q46·CSAT · Prelims 2026

Quantitative — Diophantine Digit Reversal

NumericalNumber SystemArithmetic ProblemHard

Question

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, then how many such different A's are possible? [cite: 4440, 4441, 4443, 4446, 4494]

Options

a

6

b

9

c

12

Answer
d

18

Explanation

Let the two-digit number A be represented as 10x + y, where x and y are distinct single-digit integers (1 \le x \le 9 and 0 \le y \le 9).

Formulate the algebraic expression: Reversing the digits gives B = 10y + x. The problem states that A - B is a positive multiple of 27.

A - B = (10x + y) - (10y + x) = 9(x - y)

Incorporate the condition: Set 9(x - y) equal to a multiple of 27 (27k, where k \ge 1):

9(x - y) = 27k \implies x - y = 3k

Analyze valid digit pairs based on ⟨MATH⟩k⟨/MATH⟩:
Case 1: ⟨MATH⟩k = 1 \implies x - y = 3⟨/MATH⟩. The valid (x, y) coordinate pairs are:

\{(4,1), (5,2), (6,3), (7,4), (8,5), (9,6), (3,0)\} \rightarrow 7 possibilities

Case 2: ⟨MATH⟩k = 2 \implies x - y = 6⟨/MATH⟩. The valid (x, y) coordinate pairs are:

\{(7,1), (8,2), (9,3), (6,0)\} \rightarrow 4 possibilities

Case 3: ⟨MATH⟩k = 3 \implies x - y = 9⟨/MATH⟩. The valid (x, y) coordinate pairs are:

\{(9,0)\} \rightarrow 1 possibility

Calculate the total count: Sum the valid cases: 7 + 4 + 1 = 12 total possible values for A.
The absolute difference between any two-digit number and its reverse is always equal to 9 times the difference between its digits (9|x - y|).

Answer: (c).

Question details

Year

2026

Paper

CSAT

Question

Q46

Section

Quantitative Aptitude

Sub-topic

Number System

Type

Arithmetic Problem

Difficulty

Hard

Source hint

Number theory — algebraic digit transformations

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