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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