Question
Find the least 4-digit number which, when divided by
21, 28, and 12, leaves a remainder of 3 in each case.Solution
Prime factorization of 21 = 3 × 7
Prime factorization of 28 = 2² × 7
Prime factorization of 12 = 2² × 3 LCM of (21, 28, and 12) = 2² × 3 × 7 = 84 Least 4-digit number divisible by 84 = 1,008 So, required number = 1008 + 3 = 1,011 Hence, option A.
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