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    Question

    What is the largest sum of m and n such that

    тАШ4m932n6тАЩ is divisible by 12?
    A 23 Correct Answer Incorrect Answer
    B 15 Correct Answer Incorrect Answer
    C 20 Correct Answer Incorrect Answer
    D 12 Correct Answer Incorrect Answer

    Solution

    ATQ,

    Divisible by 12 тЗТ divisible by 3 and 4

    Last two digits: тАШn6тАЩ divisible by 4 тЖТ n = 0, 2, 4, 6, 8

    Take maximum n = 8

    Sum = 4 + m + 9 + 3 + 2 + 8 + 6 = 32 + m

    32 + m divisible by 3 тЖТ m = 1, 4, 7

    Max m = 7

    тЗТ max (m + n) = 7 + 8 = 15

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