Question
Find the smallest integer greater than 100 which leaves a remainder 1 when divided by each of 2, 3 and 5.
Solution
If N leaves remainder 1 when divided by 2, 3, 5, then N − 1 is divisible by all 2, 3, 5. LCM(2, 3, 5) = 30 So N − 1 = 30k ⇒ N = 30k + 1 We need N > 100: 30k + 1 > 100 ⇒ 30k > 99 ⇒ k > 3.3 ⇒ k = 4 N = 30×4 + 1 = 121
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