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    Question

    If 'x' is the smallest positive integer divisible by 9,

    15, and 28, then determine the second smallest positive integer that is divisible by all three of these numbers.
    A 2400 Correct Answer Incorrect Answer
    B 2520 Correct Answer Incorrect Answer
    C 3420 Correct Answer Incorrect Answer
    D 3260 Correct Answer Incorrect Answer

    Solution

    Lowest positive integer divisible by 9, 15 and 28 = LCM (9, 15, 28) So, LCM of (9, 15, and 28) = 22 X 32 X 51 X 71 = 1260 Therefore, the next larger number which is divisible by all the three given numbers = 2 X 1260 = 2520

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