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I tried to explain that A also is sufficient. ( pitts20042006 corrected me - so I am modifying my post )
Answer - B is sufficent.
Choice A - if square N is integer then sqrt of N may or may not be integer.
Choice B - if sqrt of N is integer then N has to be integer. If we consider 2 or 3 as integers(n=2 or N=3) then their sqrt is not an integer. That does not mean that the sqrt(N) if it is integer then N is not integer. B would have been wrong if the question were is sqrt(N) integer ? given N as integer.
For example sqrt(4) = 2 which an integer meaning N=4 which is definitely integer.
Last edited by anandnk on 12 Dec 2003, 20:47, edited 1 time in total.