Quick view of $2050$ .

I cheated and solved this in a second: The answer must be the largest of the given choices, $\boxed{128}$ ;)

Let $m$ be the highest power of $17$ which divides $2050!$ .

$\begin{aligned} m & = \left \lfloor \frac{2050}{17} \right \rfloor + \left \lfloor \frac{2050}{17^2} \right \rfloor \\ & = 120+7 \\ & = 127 \end{aligned}$

Therefore, the any value $n>m$ will not divide $2050!$ .

Answer = $128$

We just need to count the number of 17s in $2050!$ . We can simplify the calculation by something like this -

$\sum _{ n=1 }^{ k }{ \left\lfloor \frac { 2050 }{ { 17 }^{ n } } \right\rfloor }$

$k$ must be equal to last $n$ such that

$\left\lfloor \frac { 2050 }{ { 17 }^{ n } } \right\rfloor >0$

After performing the calculations, we get the answer as 127. If ${ 17 }^{ 127 }|2050!$ then every smaller positive integral power less than $127$ also divides $2050!$ . Hence only 128 from the list doesn't divide $2050!$ .