lf n > 0 and exactly 15 integers satisfy $(\mathrm{x}+6)(\mathrm{x}-4)(\mathrm{x}-5)(2\mathrm{x}-\mathrm{n})\le 0$ then sum of digits of the least possible value of n is

$(\mathrm{x}+6)(\mathrm{x}-4)(\mathrm{x}-5)(2\mathrm{x}-\mathrm{n})\le 0$

Hence, $\mathrm{x}\in [-6,4]\cup \left[5,\frac{\mathrm{n}}{2}\right]$

Now, there are 11 integers in [-6, 4].

So, there must be 4integers in $\left[5,\frac{\mathrm{n}}{2}\right]$

So, $\frac{\mathrm{n}}{2}\in [8,9)$

$\Rightarrow \mathrm{n}\in [16,18)$