Question
Find the smallest number greater than 100 which, when
divided by 8, 12 and 15, leaves a remainder of 5 in each case.Solution
Let the required number be N. N − 5 is divisible by 8, 12 and 15. LCM of 8, 12 and 15: 8 = 2³ 12 = 2² × 3 15 = 3 × 5 LCM = 2³ × 3 × 5 = 8 × 3 × 5 = 120 So, N − 5 = 120k ⇒ N = 120k + 5 We need N > 100. For k = 1: N = 120 × 1 + 5 = 125 (> 100)
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