if (x) is a positive integer, for how many different values of (x) is (sqrt{\frac{48}{x}}) a whole…

Explanation:

Step1: Factorize 48

$48 = 2^4\times3^1$

Step2: For $\sqrt{\frac{48}{x}}$ to be a whole number, $x$ must be a factor of 48 such that $\frac{48}{x}$ is a perfect - square.

Let $x = 2^a\times3^b$, where $0\leq a\leq4$ and $0\leq b\leq1$. Then $\frac{48}{x}=2^{4 - a}\times3^{1 - b}$. For $\frac{48}{x}$ to be a perfect - square, the exponents of its prime - factors must be even. Case 1: When $b = 1$, then $4 - a$ must be even. So $a = 0,2,4$. When $a = 0,b = 1,x = 3$; when $a = 2,b = 1,x = 12$; when $a = 4,b = 1,x = 48$. Case 2: When $b = 0$, then $4 - a$ must be even. So $a = 0,2,4$. When $a = 0,b = 0,x = 1$; when $a = 2,b = 0,x = 4$; when $a = 4,b = 0,x = 16$.

Step3: Count the number of valid $x$ values

The values of $x$ are 1, 3, 4, 12, 16, 48. So there are 6 different values of $x$.

Answer:

6