UPSC Prelims 2025 CSAT, question 18

Topic: Number system

Q18

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 $N$ ?

Solution.

Answer: (c) 1078

In each case, the divisor minus the remainder is $3 - 1 = 5 - 3 = 6 - 4 = 9 - 7 = 2$ , so $N + 2$ must be a common multiple of $3, 5, 6,$ and $9$ .
Since $\operatorname{LCM}(3, 5, 6, 9) = 90$ , we have $N = 90k - 2$ for some positive integer $k$ .
The smallest 4-digit multiple of $90$ is $90 \times 12 = 1080$ , which gives the smallest 4-digit value $N = 1080 - 2 = 1078$ .