Q28. A 4-digit number N is such that when divided by 3, 5, 6, 9 it leaves a remainder of 1, 3, 4, 7 respectively. What is the smallest value of N?

We are given the following conditions for a 4-digit number N: N ≡ 1 (mod 3) => (3 - 1) = 2 N ≡ 3 (mod 5) => (5 - 3) = 2 N ≡ 4 (mod 6) => (6 - 4) = 2 N ≡ 7 (mod 9) => (9 - 7) = 2 Since the difference between the divisor and the remainder is constant (which is 2 in this case), the number N must be of the form (LCM of divisors) × k - 2, where k is an integer. First, find the LCM of the divisors (3, 5, 6, 9): LCM(3, 5, 6, 9) = LCM(3, 5, 2×3, 3^2) To find the LCM, take the highest power of each prime factor present: Prime factors are 2, 3, 5.

Highest power of 2 is 2^1 (from 6) Highest power of 3 is 3^2 (from 9) Highest power of 5 is 5^1 (from 5) LCM = 2^1 × 3^2 × 5^1 = 2 × 9 × 5 = 90. So, N = 90k - 2. We need the smallest 4-digit number N.

This means N ≥ 1000. 90k - 2 ≥ 1000 90k ≥ 1002 k ≥ 1002 / 90 k ≥ 11.13... The smallest integer value for k that satisfies this condition is 12. Substitute k = 12 into the expression for N: N = 90 × 12 - 2 N = 1080 - 2 N = 1078.

This question tests the application of the Chinese Remainder Theorem concept (specifically, when the difference between divisor and remainder is constant) and LCM, which is a standard topic in UPSC CSAT number system.