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 β 5 is divisible by 8, 12 and 15. So N β 5 = LCM(8, 12, 15) 8 = 2Β³, 12 = 2Β² Γ 3, 15 = 3 Γ 5 LCM = 2Β³ Γ 3 Γ 5 = 8 Γ 3 Γ 5 = 120 So N β 5 = 120 β N = 125
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