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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
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.

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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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