Question
Determine the smallest 4-digit number that, when divided
by 15, 24, and 30, leaves a remainder of 5 each time.Solution
Prime factorization of 15 = 3 1 Â X 5 1 Prime factorization of 24 = 2 3 Â X 3 1 Prime factorization of 30 = 2 1 Â X 3 1 Â X 5 1 LCM of (15, 24, and 30) = 2 3 Â X 3 1 Â X 5 1 Â = 8 X 3 X 5 = 120 Least 4-digit number divisible by 120 = 1,080 So, required number = 1080 + 5 = 1,085
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