Just E

The given function f(x) is continuous at all x and so is the polynomial $\sum_{r=0}^{19} x^r$ and their graphs would merge with one another. So:- $f(x) = \sum_{r=0}^{19} x^r$ So $y=\sum_{r=0}^{19}( \int_{0}^{\infty} x^{r}e^{-x} dx)$

$y=\sum_{r=0}^{19} \Gamma(r+1)$

$y=\sum_{r=0}^{19} r!$

To compute $y \mod 100$ we need $y \mod 4$ and $y \mod 25$

For the first part it is just congruent to $(\sum_{r=0}^{3} r!)$ $\mod 4$

So the $y\equiv 1+1+2+6 \mod 4 \equiv 2\mod4$

For the 2nd part $y\mod25$ $\equiv$ $(\sum_{r=0}^{9} r!)\mod25$ .

A simple computation using $5! \equiv -5\mod25$ shows us that it is $14\mod25$

So y is $2 \mod 4$ and $14 \mod 25$ which gives us nothing but $y \equiv$ $14$ $\mod 100$

So our answer is $99+14 = 133$

Relevant wiki: - Gamma Function .

Chinese Remainder Theorem