Find the smallest 3-digit number which when divided by 6, 8 and 9 leaves the same remainder 3 in each case.

Solution

If a number N leaves remainder 3 when divided by 6, 8 and 9, then N − 3 is exactly divisible by 6, 8 and 9. So N − 3 must be a multiple of LCM(6, 8, 9). Find LCM: 6 = 2 × 3 8 = 2³ 9 = 3² LCM = 2³ × 3² = 8 × 9 = 72 So N − 3 = 72k for some integer k ⇒ N = 72k + 3 We need smallest 3-digit N (≥ 100): Try k = 1: N = 72 × 1 + 3 = 75 (2-digit) k = 2: N = 144 + 3 = 147 (3-digit, and smallest) Check: 147 ÷ 6 = 24 remainder 3 147 ÷ 8 = 18 remainder 3 147 ÷ 9 = 16 remainder 3 Answer: 147.