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    Question

    Find the least 4-digit number which, when divided by

    18, 30, and 45, leaves a remainder of 2 in each case.
    A 1,082 Correct Answer Incorrect Answer
    B 1,182 Correct Answer Incorrect Answer
    C 1,282 Correct Answer Incorrect Answer
    D 1,382 Correct Answer Incorrect Answer

    Solution

    Prime factorization of 18 = 2 × 3²
    Prime factorization of 30 = 2 × 3 × 5
    Prime factorization of 45 = 3² × 5 LCM of (18, 30, and 45) = 2 × 3² × 5 = 90 Least 4-digit number divisible by 90 = 1,080 So, required number = 1080 + 2 = 1,082 Hence, option A. 

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