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?

Each remainder is 2 less than the divisor, so N + 2 is a multiple of LCM(3, 5, 6, 9) = 90. N = 90k - 2; the smallest 4-digit value is 90 × 12 - 2 = 1078. (Answer as per the official UPSC answer key.)