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    Question

    Let ‘n’ be the smallest perfect square divisible by 22,

    26, 33 and 39. What is the remainder when ‘n’ is divided by 100?
    A 53 Correct Answer Incorrect Answer
    B 14 Correct Answer Incorrect Answer
    C 30 Correct Answer Incorrect Answer
    D 64 Correct Answer Incorrect Answer

    Solution

    ATQ,

    LCM of 22, 26, 33 and 39 = 2574

    Since, 2574 = (2 × 3 × 11 × 13 × 3)

    = 2 × 3 × 3 × 11 × 13

    To make it a perfect square, multiply by (2 × 11 × 13) = 286

    So, required number = 2574 × 286 = 736164

    736164 ÷ 100 = 7361 × 100 + 64

    So, desired remainder = 64

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