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What is the remainder when the two-digit, positive integer.

This topic has 1 expert reply and 0 member replies

What is the remainder when the two-digit, positive integer.

Post Wed Sep 20, 2017 7:24 am
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    What is the remainder when the two-digit, positive integer x is divided by 3?

    (1) The sum of the digits of x is 5
    (2) The remainder when x is divided by 9 is 5

    The OA is D.

    I got confused. I could not determine the statements alone are sufficient. Can any expert help me here?

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    Post Wed Sep 20, 2017 11:52 pm
    Vincen wrote:
    What is the remainder when the two-digit, positive integer x is divided by 3?

    (1) The sum of the digits of x is 5
    (2) The remainder when x is divided by 9 is 5

    The OA is D.

    I got confused. I could not determine the statements alone are sufficient. Can any expert help me here?
    Statement 1: The sum of the digits of x is 5.

    Divisibility rule of 3:

    If the sum of the digits of a number is divisible by 3, then the number is divisible by 3.

    However, if the sum of the digits of a number is NOT divisible by 3, then the remainder is the same when the number is divided by 3.

    Thus, the remainder when the two-digit, positive integer x is divided by 3 = Remainder when 5 is divided by 3 = 2. Sufficient.

    Statement 2: The remainder when x is divided by 9 is 5.

    The rule for divisibility of 9 is same as that for 3.

    The remainder when the two-digit, positive integer x is divided by 3 = Remainder when 5 is divided by 3 = 2. Sufficient.

    Alternatively,

    Say X = 9q + 5, where q is quotient

    Thus, X/3 = (9/3)*q + 5/3

    X/3 = 3q + 1 + 2/3

    => Remainder = 2. Sufficient.

    The correct answer: D

    Hope this helps!

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    Thanked by: Vincen

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