If n is a positive integer, is n^(1/2)/3 an integer ?

Solution

Step 1: Analyse Question Stem

• n is a positive integer.
• we need to find if\( \frac{\sqrt{n}}{3} = I\), where I is a positive integer.

o \(\frac{\sqrt{n}}{3} = I ⟹\) \(\sqrt{\frac{n}{9}} = I ⟹ n= 9*I^2 ⟹ n= (3*I)^2\)

 This means, n is square of a multiple of 3.

Step 2: Analyse Statements Independently (And eliminate options) – AD/BCE

• According to this statement: \(\sqrt{\frac{n}{3}} = m\), where m is a positive integer.

o This means, \(\frac{n}{3} = m^2\)
o Or, \(n = 3*m^2\)

 From the above, we can see that power of 3 is not even, hence n is not a perfect square.
• Therefore, we can say that \(\frac{\sqrt{n}}{3}\) is not an integer.

• From this statement, all we know is that, 14n is not a perfect square.
• However, we cannot conclude anything about n, for example, consider the following two cases:

o Case 1: if \(n=2\) then \(14n = 28\) and \(\sqrt{14n}\) is not an integer.

 Here n is not even perfect square.
o Case 2: If \(n = 9\), then \(14n = 14*3^2\) and \(\sqrt{14n}\) is not an integer.

 However, now n is a square of multiple of 3