1. Let the original length and area of cross-section of the wire be $L$ and $A$ respectively.

2. Original resistance: $R = \rho \frac{L}{A}$ where $\rho$ is the resistivity.

3. The wire is stretched, so its material remains the same and its volume stays constant: $A L = A' L'$

4. Original radius is $r$, new radius is $r/2$. So the new area becomes $A' = \pi \left(\frac{r}{2}\right)^2 = \frac{\pi r^2}{4} = \frac{A}{4}$

5. Using volume conservation: $A L = A' L'$ $A L = \frac{A}{4} L'$ $L' = 4L$

6. New resistance: $R' = \rho \frac{L'}{A'} = \rho \frac{4L}{A/4} = 16\rho \frac{L}{A}$ $R' = 16R$

7. Since the new resistance is given as $xR$, $xR = 16R$ $x = 16$

Therefore, the required integer is: $\boxed{16}$