Question
Find the smallest 4-digit number which leaves a remainder 5 when divided by each of 6, 8 and 9.
Solution
If N leaves remainder 5 when divided by 6, 8 and 9, then N − 5 is divisible by all three. LCM(6, 8, 9): 6 = 2 × 3 8 = 2³ 9 = 3² LCM = 2³ × 3² = 8 × 9 = 72 So N − 5 = 72k N = 72k + 5 We want smallest 4-digit N ≥ 1000: 72k + 5 ≥ 1000 ⇒ 72k ≥ 995 ⇒ k ≥ 995/72 ≈ 13.8 So k = 14 N = 72 × 14 + 5 = 1,008 + 5 = 1,013 Answer: 1,013.
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