Question

    If the number 8357pt is divisible by 15, how many values

    can 'p' take?
    A 12 Correct Answer Incorrect Answer
    B 10 Correct Answer Incorrect Answer
    C 6 Correct Answer Incorrect Answer
    D 2 Correct Answer Incorrect Answer

    Solution

    ATQ,

    Divisible by 15 ⇒ divisible by both 5 and 3

    Divisible by 5 ⇒ t = 0 or 5

    Divisible by 3 ⇒ sum divisible by 3

    Case 1: t = 0

    Sum = 8 + 3 + 5 + 7 + p + 0 = p + 23

    Divisible by 3 ⇒ p = 1, 4, 7

    Case 2: t = 5

    Sum = 8 + 3 + 5 + 7 + p + 5 = p + 28

    Divisible by 3 ⇒ p = 2, 5, 8

    Total values of p: 3 + 3 = 6 values

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