Question

    Find the difference between minimum and maximum value

    of 'j' such that '7j5302' is always divisible by 3.
    A 7 Correct Answer Incorrect Answer
    B 9 Correct Answer Incorrect Answer
    C 6 Correct Answer Incorrect Answer
    D 5 Correct Answer Incorrect Answer

    Solution

    A number is divisible by 3 when the sum of its digits is divisible by 3.
    Sum of digits of '7j5302' = (7 + j + 5 + 3 + 0 + 2) = j + 17.
    So, j + 17 should be divisible by 3.
    Possible values of 'j' = 1, 4, 7
    Minimum value of 'j' = 1
    Maximum value of 'j' = 7
    Required difference = 7 – 1 = 6

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