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Q8·CSAT · Prelims 2025

4-digit number with remainder conditions

NumericalRemainders & DivisibilityNumber theoryHard

Question

A 4-digit number N is such that when divided by 3, 5, 6, 9 leaves a remainder 1, 3, 4, 7 respectively. What is the smallest value of N?

Options

a

1068

b

1072

c

1078

Answer
d

1082

Explanation

The difference between each divisor and its corresponding remainder is constant: 3-1 = 2, 5-3 = 2, 6-4 = 2, and 9-7 = 2. When this difference (d) is uniform, the required number takes the form LCM(divisors) * k - d.

The LCM of 3, 5, 6, and 9 is 90. The number must be of the form 90k - 2. We need the smallest 4-digit number. The smallest 4-digit multiple of 90 is 90 * 12 = 1080. Subtracting the constant difference: 1080 - 2 = 1078.

If division by a, b, c leaves remainders x, y, z and a-x = b-y = c-z = d, the number is always a multiple of LCM(a,b,c) - d.

Answer: (c).

Question details

Year

2025

Paper

CSAT

Question

Q8

Section

Numerical Ability

Sub-topic

Remainders & Divisibility

Type

Number theory

Difficulty

Hard

Source hint

Number theory

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