UPSC CSE Prelims 2025
The problem requires finding the smallest 4-digit number N that satisfies a set of remainder conditions:
Observe the difference between each divisor and its corresponding remainder:
Since this difference is constant (2), it implies that N + 2 is perfectly divisible by 3, 5, 6, and 9. Therefore, N + 2 must be a multiple of the Least Common Multiple (LCM) of these divisors.
Calculate the LCM of 3, 5, 6, and 9:
The LCM is obtained by taking the highest power of each prime factor present: LCM(3, 5, 6, 9) = 2¹ × 3² × 5¹ = 2 × 9 × 5 = 90.
Thus, N + 2 must be a multiple of 90. We can write this as N + 2 = 90k, where k is a positive integer. Rearranging for N, we get N = 90k - 2.
To find the smallest 4-digit number N, we set N ≥ 1000:
90k - 2 ≥ 1000
90k ≥ 1002
k ≥ 1002 / 90
k ≥ 11.133...
Since k must be an integer, the smallest integer value for k that satisfies the condition is 12.
Substitute k = 12 into the equation for N:
N = 90 × 12 - 2
N = 1080 - 2
N = 1078
The number 1078 is a 4-digit number and satisfies all the given remainder conditions.
The following options do not satisfy all the given remainder conditions:
Option (1) 1068:
Therefore, 1068 is not the correct number.
Option (2) 1072:
Therefore, 1072 is not the correct number.
Option (4) 1082:
Therefore, 1082 is not the correct number.