Vedadots
Q3435/80Q36
Q35·CSAT · Prelims 2024

Finding the smallest number satisfying multiple remainder conditions

NumericalLCM & HCFFactual singleMedium

Question

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?

Options

a

2

b

4

c

6

d

8

Answer

Explanation

Let X = 10a + b, where a and b are single-digit integers (a \ne 0). Interchanging the digits gives Y = 10b + a (b \ne 0 for Y to remain a valid two-digit number) [cite: 1567, 1614]. The sum is: X + Y = (10a + b) + (10b + a) = 11(a + b) .

We are given that (X + Y) is the greatest two-digit number, which is 99[cite: 1568, 1614]. 11(a + b) = 99 \implies a + b = 9.

Identify all valid single-digit pairs (a, b) such that a \ge 1 and b \ge 1: (1, 8), (2, 7), (3, 6), (4, 5), (5, 4), (6, 3), (7, 2), (8, 1). There are exactly 8 valid configurations for X.

The sum of any two-digit number and its digit-reversed counterpart is always a multiple of 11, specifically 11(a+b).

Answer: (d).

Question details

Year

2024

Paper

CSAT

Question

Q35

Section

Numerical Ability

Sub-topic

LCM & HCF

Type

Factual single

Difficulty

Medium

Source hint

LCM and remainder application

Same sub-topic — other years

LCM & HCF has appeared in multiple CSAT papers:

See all questions on LCM & HCF

Browse every tagged question across all years

Explore →