📢 Too many exams? Don’t know which one suits you best? Book Your Free Expert 👉 call Now!


    Question

    If 'x' is the lowest positive integer divisible by 10,

    18 and 21, then find the second smallest positive integer which is divisible by all the three given numbers?
    A 630 Correct Answer Incorrect Answer
    B 1260 Correct Answer Incorrect Answer
    C 1890 Correct Answer Incorrect Answer
    D 2520 Correct Answer Incorrect Answer

    Solution

    Lowest positive integer divisible by 10, 18 and 21 = LCM (10, 18, 21) So, LCM of (10, 18, and 21) = 2¹ × 3² × 5¹ × 7¹ = 630 Therefore, the next larger number which is divisible by all the three given numbers = 2 × 630 = 1260

    Practice Next
    ask-question