Is x the square of an Integer?

Hi,

Questions asks whether x=I^2 where I is a Integer.

from St 1 we have that x= 12K+6 or x= 6(2K+1)
Now when can 6*(2k+1) will be square??
2k+1 has to equal to 6 or 24 (6*2*2) or 54 (6*3*3) or 96 (6*4*4) to make 6*(2k+1) as square of integer.

But if 6=2k+1 then k =5/2 which is not an integer and is the case for all the values as well.

Hence St 1 is sufficient and option B,C and E ruled out

St 2 we have x=3q+9 or x=3 (q+3) now Q is a positive integer and for values of q =9, x=36 ie. square of an integer and if q= 1,x=12 ie. not a square of an integer.

Ans A

Question: Is x a perfect square?

For x to be a perfect square, its each prime factor should have an even exponent in the prime factorisation.

(1) x = 12k + 6, where k is a positive integer
x = 6(2k + 1) = 2 * 3 * (2k+1)
(2k+1) is certainly odd. Hence the exponent of 2 in its prime factorization is 1 only.
Hence x cannot be prefect square. Sufficient alone.

(2) x = 3q + 9, where q is a positive integer
x = 3(q + 3)
q could be 1 in which case x is not a perfect square or it could take a value such that we get another 3. Say q could be 24 so that x = 3*27 = 3^4 (and hence 3 has an even exponent)
So here, x is a perfect square. Not sufficient alone

Answer (A)

Check this video on perfect squares: https://anaprep.com/number-properties-f ... ct-square/