Let `S = sum_(n =1)^(99) (5^(100))/((25)^(n) + 5^(100))`. Find [S]. Where [y] denotes largest integer less than or equal to y.

Answer

To solve the problem, we need to evaluate the sum

\[ S = \sum_{n=1}^{99} \frac{5^{100}}{25^n + 5^{100}}. \]

### Step 1: Rewrite the terms in the summation
We can rewrite \(25^n\) as \((5^2)^n = 5^{2n}\). Thus, we can express the sum as:

\[ S = \sum_{n=1}^{99} \frac{5^{100}}{5^{2n} + 5^{100}}. \]

### Step 2: Simplify each term
Now, we can factor out \(5^{100}\) from the denominator:

\[ S = \sum_{n=1}^{99} \frac{5^{100}}{5^{100}(5^{2n-100} + 1)} = \sum_{n=1}^{99} \frac{1}{5^{2n-100} + 1}. \]

### Step 3: Pair the terms
Notice that for each \(n\), we can pair terms \(t_n\) and \(t_{100-n}\):

\[ t_n = \frac{1}{5^{2n-100} + 1} \quad \text{and} \quad t_{100-n} = \frac{1}{5^{200-2n-100} + 1} = \frac{1}{5^{100-2n} + 1}. \]

### Step 4: Add the pairs
Now, let's add \(t_n\) and \(t_{100-n}\):

\[ t_n + t_{100-n} = \frac{1}{5^{2n-100} + 1} + \frac{1}{5^{100-2n} + 1}. \]

Finding a common denominator:

\[ = \frac{(5^{100-2n} + 1) + (5^{2n-100} + 1)}{(5^{2n-100} + 1)(5^{100-2n} + 1)}. \]

This simplifies to:

\[ = \frac{5^{100-2n} + 5^{2n-100} + 2}{(5^{2n-100} + 1)(5^{100-2n} + 1)}. \]

However, we notice that \(t_n + t_{100-n} = 1\) because:

\[ t_n + t_{100-n} = \frac{1}{5^{2n-100} + 1} + \frac{1}{5^{100-2n} + 1} = 1. \]

### Step 5: Count the pairs
Since \(n\) runs from 1 to 99, we have 49 pairs (from \(n=1\) to \(n=49\)) and one middle term when \(n=50\):

\[ t_{50} = \frac{1}{5^{100-100} + 1} = \frac{1}{1 + 1} = \frac{1}{2}. \]

### Step 6: Calculate the total sum
Thus, the total sum \(S\) can be calculated as:

\[ S = 49 \cdot 1 + \frac{1}{2} = 49 + 0.5 = 49.5. \]

### Step 7: Find the greatest integer less than or equal to \(S\)
Finally, we need to find \([S]\):

\[ [S] = [49.5] = 49. \]

### Final Answer
Thus, the answer is:

\[ \boxed{49}. \]