Q. 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?

Explanation : 

1. Restate the three active conditions (the “÷3 ⇒1” is automatically met once the others hold):
• On dividing by 6, remainder = 4
• On dividing by 9, remainder = 7
• On dividing by 5, remainder = 3

2. Since lcm(6, 9, 5) = 90, any solution N will look like
N = X + 90·k,
where X is some number from 0 to 89 that fits all three remainders, and k is a non-negative integer.

3. Find all X in [0…89] with remainder 4 when divided by 6:
Start at 4, then keep adding 6 →
4, 10, 16, 22, 28, 34, 40, 46, 52, 58, 64, 70, 76, 82, 88 (15 candidates)

4. From these, pick those that leave remainder 7 when divided by 9:
Compute each mod 9:
– 4 →4
– 10→1
– 16→7 ✓
– 22→4
– 28→1
– 34→7 ✓
– 40→4
– 46→1
– 52→7 ✓
– 58→4
– 64→1
– 70→7 ✓
– 76→4
– 82→1
– 88→7 ✓
So the survivors are X = 16, 34, 52, 70, 88.

5. From those five, pick the one with remainder 3 on division by 5:
– 16→1, 34→4, 52→2, 70→0, 88→3 ✓
So X = 88 is the unique fit in [0…89].

6. General solution:
N = 88 + 90·k

7. We want the smallest 4-digit N ≥ 1000:
88 + 90·k ≥ 1000
90·k ≥ 912
k ≥ 912/90 = 10.13… ⇒ k = 11

⇒ N = 88 + 90·11 = 88 + 990 = 1078.

Answer: 1078 (option c)