Find the ordered triple of natural numbers $(x,y,z)$ such that $x \le y \le z$ and $x^x+y^y+z^z = 3382.$

[i]Proposed by Evan Fang

1. **Determine the upper bound for \( z \):**
   - We need to find the maximum value of \( z \) such that \( z^z \leq 3382 \).
   - Calculate \( z^z \) for small values of \( z \):
     \[
     \begin{aligned}
     1^1 &= 1, \\
     2^2 &= 4, \\
     3^3 &= 27, \\
     4^4 &= 256, \\
     5^5 &= 3125, \\
     6^6 &= 46656.
     \end{aligned}
     \]
   - Since \( 5^5 = 3125 \) and \( 6^6 = 46656 \), we conclude that \( z \leq 5 \).

2. **Check the case \( z = 5 \):**
   - If \( z = 5 \), then \( 5^5 = 3125 \).
   - We need \( x^x + y^y = 3382 - 3125 = 257 \).

3. **Find pairs \((x, y)\) such that \( x^x + y^y = 257 \):**
   - Calculate \( x^x \) and \( y^y \) for small values of \( x \) and \( y \):
     \[
     \begin{aligned}
     1^1 &= 1, \\
     2^2 &= 4, \\
     3^3 &= 27, \\
     4^4 &= 256.
     \end{aligned}
     \]
   - Check combinations:
     \[
     \begin{aligned}
     1^1 + 4^4 &= 1 + 256 = 257, \\
     2^2 + 3^3 &= 4 + 27 = 31, \\
     3^3 + 3^3 &= 27 + 27 = 54.
     \end{aligned}
     \]
   - The only combination that works is \( (x, y) = (1, 4) \).

4. **Verify the solution:**
   - We have \( x = 1 \), \( y = 4 \), and \( z = 5 \).
   - Check the sum:
     \[
     1^1 + 4^4 + 5^5 = 1 + 256 + 3125 = 3382.
     \]
   - The conditions \( x \leq y \leq z \) are satisfied.

Conclusion:
The ordered triple \((x, y, z)\) that satisfies the given conditions is \((1, 4, 5)\).

The final answer is \(\boxed{(1, 4, 5)}\).