This looks easy to me...

To get the value of $x$ ,we factorize $S$ into $5^{200} \times ( 1 \times 2 \times 3 ...... \times 200)$ because there are 200 terms and each of them has a common factor of 5. Now,we calculate the number of 2's and 5's in the prime factorization of the product. Using de Polignac's ( or Legendre's ) formula,we find that there are 197 2's and 49 5's.However,there are an additional 200 5's in the term $5^{200}$ , so there are a total of 197 2's and 249 5's.Pairing a 2 with a 5 gives 10 which adds an extra trailing zero,so there are 197 zeros at the end of $S$ .So, $x = 197$ .Using the same method of factorization,we find that $T = 10^{100} \times ( 1 \times 2 \times 3 \times 4 ...... \times 100)$ There are 24 zeros at the end of the product and 100 zeros in the term $10^{100}$ .So,there are 124 zeros at the end of $T$ and $y = 124$ .Because 197 is a prime number, the fraction $\frac{x}{y}$ is irreducible,so $a = x$ and $b = y$ .Then, $\frac{ab}{31} = \frac{197 \times 124}{31} = \frac{197 \times 4 \times 31}{31} = 197 \times 4 = \boxed{788}$