- A. 1068
- B. 1072
- C. 1078
- D. 1082
Answer: C
Explanation
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.