Find the smallest positive value of $36^k - 5^m$, where $k$ and $m$ are positive integers.
1. **Understanding the problem**: We need to find the smallest positive value of \(36^k - 5^m\) where \(k\) and \(m\) are positive integers.

2. **Modulo 5 analysis**: 
   - Note that \(36 \equiv 1 \pmod{5}\). Therefore, \(36^k \equiv 1^k \equiv 1 \pmod{5}\).
   - This implies \(36^k - 5^m \equiv 1 - 0 \equiv 1 \pmod{5}\).

3. **Possible values**: 
   - From the above, \(36^k - 5^m\) must be congruent to 1 modulo 5. Hence, the possible values are of the form \(5n + 1\) for some integer \(n\).

4. **Modulo 6 analysis**:
   - Note that \(36 \equiv 0 \pmod{6}\). Therefore, \(36^k \equiv 0 \pmod{6}\).
   - Also, \(5 \equiv -1 \pmod{6}\). Therefore, \(5^m \equiv (-1)^m \pmod{6}\).
   - This implies \(36^k - 5^m \equiv 0 - (-1)^m \equiv (-1)^{m+1} \pmod{6}\).

5. **Checking smaller values**:
   - We need to check if \(36^k - 5^m = 1\) is possible.
   - Assume for contradiction that \(36^k - 5^m = 1\). This rearranges to \((6^k-1)(6^k+1) = 5^m\).

6. **Analyzing the factors**:
   - Since \(m \geq 1\), \(6^k-1\) and \(6^k+1\) must both be powers of 5.
   - Let \(6^k - 1 = 5^a\) and \(6^k + 1 = 5^b\) where \(a\) and \(b\) are integers.
   - This implies \(5^b - 5^a = 2\).

7. **Contradiction**:
   - The smallest difference between two powers of 5 is \(5 - 1 = 4\), which is greater than 2.
   - By induction, the difference between two powers of 5 only increases as the powers increase.
   - Therefore, there are no two powers of 5 that differ by 2, leading to a contradiction.

8. **Conclusion**:
   - Since \(1\) is not possible, we need to check the next smallest value.
   - The next smallest value that is congruent to 1 modulo 5 and not divisible by 6 is \(11\).

The final answer is \(\boxed{11}\).