We have shortened the usual notation indicating with a sub-index the number of times that a digit is conseutively repeated. For example, $1119900009$ is denoted $1_3 9_2 0_4 9_1$.

Find $(x, y, z)$ if  $2_x 3_y 5_z + 3_z 5_x 2_y = 5_3 7_2 8_3 5_1 7_3$
1. **Rewrite the given equation in expanded form:**
   \[
   2_x 3_y 5_z + 3_z 5_x 2_y = 5_3 7_2 8_3 5_1 7_3
   \]
   This means we need to find \(x\), \(y\), and \(z\) such that the sum of the two sequences on the left equals the sequence on the right.

2. **Analyze the right-hand side:**
   The sequence \(5_3 7_2 8_3 5_1 7_3\) translates to:
   \[
   555 77 888 5 777
   \]
   This sequence has a total of 12 digits.

3. **Determine the total number of digits:**
   Since both summands on the left must also have 12 digits in total, we have:
   \[
   x + y + z = 12
   \]

4. **Analyze the addition without carry:**
   Since there is no carry in the addition, each digit in the result must be the sum of corresponding digits from the two sequences. 

5. **Identify the constraints from the digits:**
   - The digit 7 at the end occurs if we add 5 and 2. This implies:
     \[
     \min(y, z) = 3
     \]
   - The cross sum of the digits must add up to the sum of the digits in the result:
     \[
     2x + 3y + 5z + 3z + 5x + 2y = 5 \cdot 3 + 7 \cdot 2 + 8 \cdot 3 + 5 \cdot 1 + 7 \cdot 3
     \]
     Simplifying the right-hand side:
     \[
     2x + 3y + 5z + 3z + 5x + 2y = 15 + 14 + 24 + 5 + 21 = 79
     \]
     Combining like terms on the left-hand side:
     \[
     7x + 5y + 8z = 79
     \]

6. **Solve the system of equations:**
   We have two equations:
   \[
   x + y + z = 12
   \]
   \[
   7x + 5y + 8z = 79
   \]

7. **Substitute and solve:**
   From the first equation, express \(y\) in terms of \(x\) and \(z\):
   \[
   y = 12 - x - z
   \]
   Substitute \(y\) into the second equation:
   \[
   7x + 5(12 - x - z) + 8z = 79
   \]
   Simplify:
   \[
   7x + 60 - 5x - 5z + 8z = 79
   \]
   \[
   2x + 3z = 19
   \]

8. **Check possible values for \(z\):**
   - If \(z = 3\):
     \[
     2x + 3(3) = 19
     \]
     \[
     2x + 9 = 19
     \]
     \[
     2x = 10
     \]
     \[
     x = 5
     \]
     Then:
     \[
     y = 12 - 5 - 3 = 4
     \]
     This gives \(x = 5\), \(y = 4\), \(z = 3\).

   - If \(y = 3\):
     \[
     x + z = 9
     \]
     This contradicts \(\min(y, z) = 3\) since \(z\) would have to be less than 3.

Thus, the only valid solution is \(x = 5\), \(y = 4\), \(z = 3\).

The final answer is \( \boxed{ (x, y, z) = (5, 4, 3) } \).