If `f (x) + f (1 - x)` is equal to 10 for all real numbers x then `f ((1)/(100)) + f ((2)/(100)) + f ((3)/( 100))+...+f ((99)/(100))` equals

Answer

To solve the problem, we need to find the value of the sum \( S = f\left(\frac{1}{100}\right) + f\left(\frac{2}{100}\right) + f\left(\frac{3}{100}\right) + \ldots + f\left(\frac{99}{100}\right) \) given that \( f(x) + f(1 - x) = 10 \) for all real numbers \( x \).

### Step-by-Step Solution:

1. **Understanding the Function Property**:
We know that for any \( x \), \( f(x) + f(1 - x) = 10 \). This means that if we take any value \( x \) and its complement \( 1 - x \), their function values add up to 10.

2. **Pairing Terms in the Sum**:
We can pair the terms in the sum \( S \):
- Pair \( f\left(\frac{1}{100}\right) \) with \( f\left(\frac{99}{100}\right) \)
- Pair \( f\left(\frac{2}{100}\right) \) with \( f\left(\frac{98}{100}\right) \)
- Continue this pattern until we pair \( f\left(\frac{49}{100}\right) \) with \( f\left(\frac{51}{100}\right) \)
- Finally, we have \( f\left(\frac{50}{100}\right) \) which is unpaired.

3. **Calculating the Pairs**:
Each of these pairs sums to 10:
- \( f\left(\frac{1}{100}\right) + f\left(\frac{99}{100}\right) = 10 \)
- \( f\left(\frac{2}{100}\right) + f\left(\frac{98}{100}\right) = 10 \)
- ...
- \( f\left(\frac{49}{100}\right) + f\left(\frac{51}{100}\right) = 10 \)

There are 49 pairs, and each pair contributes 10 to the sum.

4. **Calculating the Contribution of the Pairs**:
The contribution from the 49 pairs is:
\[ 49 \times 10 = 490 \]

5. **Considering the Middle Term**:
The middle term is \( f\left(\frac{50}{100}\right) \) or \( f\left(\frac{1}{2}\right) \). Using the property:
\[ f\left(\frac{1}{2}\right) + f\left(\frac{1}{2}\right) = 10 \]
This simplifies to:
\[ 2f\left(\frac{1}{2}\right) = 10 \implies f\left(\frac{1}{2}\right) = 5 \]

6. **Final Calculation of the Sum**:
Now, adding the contributions from the pairs and the middle term:
\[ S = 490 + 5 = 495 \]

### Final Answer:
\[ S = 495 \]