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is root(n) an integer?

This topic has 1 expert reply and 0 member replies

is root(n) an integer?

Post Tue Sep 12, 2017 10:51 am
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    If n=s^at^b, where a, b, s and t are integers, is root(n) an integer?

    1) a+b is an even number
    2) a is an even number

    E is the OA.

    I got confused with this DS question. Experts please explain it to me.

    Need free GMAT or MBA advice from an expert? Register for Beat The GMAT now and post your question in these forums!
    Post Tue Sep 19, 2017 4:58 am
    Vincen wrote:
    If n=s^at^b, where a, b, s and t are integers, is root(n) an integer?

    1) a+b is an even number
    2) a is an even number

    E is the OA.

    I got confused with this DS question. Experts please explain it to me.
    We have n = s^at^b, where a, b, s and t are integers.

    We have to determine whether √n is an integer.

    √n = √(s^at^b)

    √n = s^(a/2).t^(b/2)

    Statement 1: a + b is an even number.

    Case 1: If a and b both are even, and s and t both are non-negative integers, then √n is an integer.

    Example: s = 2, t = 3, a = 2 and b = 4

    Thus, √n = s^(a/2).t^(b/2) = 2^(2/2).3^(4/2) = 2^1.3^2 = 2.9 = 18 (Integer)

    Case 2: If at least one of a and b is a negative integer, then √n is NOT an integer.

    Example: s = 2, t = 3, a = -2 and b = 4

    Thus, √n = s^(a/2).t^(b/2) = (2)^(-2/2).3^(4/2) = (2)^(-1).3^2 = 9/2 (Not an integer)

    Statement 2: a is an even number.

    Both the cases discussed above are applicable here too, thus even after combining the statements, we can ascertain for sure that √n is an integer.

    The correct answer: E

    Hope this helps!

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    Thanked by: Vincen

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