Let $x_{1}$ be a positive real number and for every integer $n \geq 1$ let $x_{n+1} = 1 + x_{1}x_{2}\ldots x_{n-1}x_{n}$. 
If $x_{5} = 43$, what is the sum of digits of the largest prime factors of $x_{6}$?
1. Given the sequence defined by \( x_{n+1} = 1 + x_1 x_2 \ldots x_n \) for \( n \geq 1 \), and \( x_5 = 43 \), we need to find the sum of the digits of the largest prime factor of \( x_6 \).

2. From the given information, we know:
   \[
   x_5 = 43
   \]
   Therefore, by the definition of the sequence:
   \[
   x_5 = 1 + x_1 x_2 x_3 x_4
   \]
   This implies:
   \[
   43 = 1 + x_1 x_2 x_3 x_4 \implies x_1 x_2 x_3 x_4 = 42
   \]

3. Next, we need to find \( x_6 \):
   \[
   x_6 = 1 + x_1 x_2 x_3 x_4 x_5
   \]
   Substituting the known values:
   \[
   x_6 = 1 + 42 \cdot 43
   \]
   Calculate the product:
   \[
   42 \cdot 43 = 1806
   \]
   Therefore:
   \[
   x_6 = 1 + 1806 = 1807
   \]

4. Now, we need to find the prime factors of 1807. We start by checking divisibility by smaller prime numbers:
   - 1807 is odd, so it is not divisible by 2.
   - The sum of the digits of 1807 is \(1 + 8 + 0 + 7 = 16\), which is not divisible by 3, so 1807 is not divisible by 3.
   - 1807 does not end in 0 or 5, so it is not divisible by 5.
   - Check divisibility by 7:
     \[
     1807 \div 7 \approx 258.14 \quad \text{(not an integer, so not divisible by 7)}
     \]
   - Check divisibility by 11:
     \[
     1807 \div 11 \approx 164.27 \quad \text{(not an integer, so not divisible by 11)}
     \]
   - Check divisibility by 13:
     \[
     1807 \div 13 = 139 \quad \text{(exact division, so 1807 = 13 \cdot 139)}
     \]

5. The prime factors of 1807 are 13 and 139. The largest prime factor is 139.

6. Finally, we find the sum of the digits of 139:
   \[
   1 + 3 + 9 = 13
   \]

The final answer is \(\boxed{13}\).