Question
Find the smallest number which when divided by 8, 12 and 15 leaves a remainder of 5 in each case.
Solution
Let the required number = N. Then N − 5 is divisible by 8, 12 and 15. So N − 5 = LCM(8, 12, 15) 8 = 2³, 12 = 2² × 3, 15 = 3 × 5 LCM = 2³ × 3 × 5 = 8 × 3 × 5 = 120 So N − 5 = 120 → N = 125
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