The positive integers $v, w, x, y$, and $z$ satisfy the equation \[v + \frac{1}{w + \frac{1}{x + \frac{1}{y + \frac{1}{z}}}} = \frac{222}{155}.\] Compute $10^4 v + 10^3 w + 10^2 x + 10 y + z$.
1. We start with the given equation:
   \[
   v + \frac{1}{w + \frac{1}{x + \frac{1}{y + \frac{1}{z}}}} = \frac{222}{155}
   \]
   Since \(v\) is a positive integer and \(1 < \frac{222}{155} < 2\), it follows that \(v = 1\).

2. Subtracting \(v = 1\) from both sides, we get:
   \[
   \frac{1}{w + \frac{1}{x + \frac{1}{y + \frac{1}{z}}}} = \frac{222}{155} - 1 = \frac{222}{155} - \frac{155}{155} = \frac{67}{155}
   \]

3. Taking the reciprocal of both sides, we have:
   \[
   w + \frac{1}{x + \frac{1}{y + \frac{1}{z}}} = \frac{155}{67}
   \]
   Since \(w\) is a positive integer and \(2 < \frac{155}{67} < 3\), it follows that \(w = 2\).

4. Subtracting \(w = 2\) from both sides, we get:
   \[
   \frac{1}{x + \frac{1}{y + \frac{1}{z}}} = \frac{155}{67} - 2 = \frac{155}{67} - \frac{134}{67} = \frac{21}{67}
   \]

5. Taking the reciprocal of both sides, we have:
   \[
   x + \frac{1}{y + \frac{1}{z}} = \frac{67}{21}
   \]
   Since \(x\) is a positive integer and \(3 < \frac{67}{21} < 4\), it follows that \(x = 3\).

6. Subtracting \(x = 3\) from both sides, we get:
   \[
   \frac{1}{y + \frac{1}{z}} = \frac{67}{21} - 3 = \frac{67}{21} - \frac{63}{21} = \frac{4}{21}
   \]

7. Taking the reciprocal of both sides, we have:
   \[
   y + \frac{1}{z} = \frac{21}{4}
   \]
   Since \(y\) is a positive integer and \(5 < \frac{21}{4} < 6\), it follows that \(y = 5\).

8. Subtracting \(y = 5\) from both sides, we get:
   \[
   \frac{1}{z} = \frac{21}{4} - 5 = \frac{21}{4} - \frac{20}{4} = \frac{1}{4}
   \]

9. Taking the reciprocal of both sides, we have:
   \[
   z = 4
   \]

10. Finally, we compute \(10^4 v + 10^3 w + 10^2 x + 10 y + z\):
    \[
    10^4 \cdot 1 + 10^3 \cdot 2 + 10^2 \cdot 3 + 10 \cdot 5 + 4 = 10000 + 2000 + 300 + 50 + 4 = 12354
    \]

The final answer is \(\boxed{12354}\).