DS- OG11 # 146

Is n an integer?

I. n squared is an integer.
II. Square root of n is an integer.

From (I) 3 squared is an integer --> n is an integer however, sqrt(2) squared is also an integer but sqrt(2) is not. Hence answer is B,C or E.

From (II) implies sqrt(n) = m where m is an integer. Squaring both sides we get n = m*m where m is an integer and hence n MUST be integer.

Hence answer is B.

Calista.

OK so I am really leaning towards qa is D. However, I dont see the logic as to why the statement 1 is insifficient.

I am plugging in answers 1,2,3,4,etc. Fractions will not work here therefore qa is imo D, which is wrong but why?

simba12123, for statement1, let n = sqrt(n), n=2, 3, 5, 6, 7... non-perfect squared numbers.

n=sqrt(2), not an integer. n^2 is an integer.
n=sqrt(4), an integer. n^2 is an integer.

therefore, statement 1 is NS.

simba12123 wrote:OK so I am really leaning towards qa is D. However, I dont see the logic as to why the statement 1 is insifficient.

I am plugging in answers 1,2,3,4,etc. Fractions will not work here therefore qa is imo D, which is wrong but why?

Simba this is what first statement tells you:

N^2 = k OR N = Square Root ( k ) , where k is integer

so you should give values to k and check for N

for k=1 N=1 - you are good
for k=2 N=1.4 - STOP!

INSUF