Question
Find the least 4-digit number which is when divided by
21, 28, and 36 leaves a remainder of 11 in each case.Solution
Prime factorization of 21 = 3 × 7 Prime factorization of 28 = 2² × 7 Prime factorization of 36 = 2² × 3² LCM of (21, 28, 36) = 2² × 3² × 7 = 252 Least 4-digit number divisible by 252 = 1,008 So, required number = 1008 + 11 = 1,019
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