For how many values of $k$ is $12^{12}$ the least common multiple of the positive integers $6^6, 8^8,$ and $k$?
1. We start by noting that the least common multiple (LCM) of the numbers \(6^6\), \(8^8\), and \(k\) must be \(12^{12}\). We need to find the values of \(k\) such that this condition holds.

2. First, we express each number in terms of its prime factorization:
   \[
   6^6 = (2 \cdot 3)^6 = 2^6 \cdot 3^6
   \]
   \[
   8^8 = (2^3)^8 = 2^{24}
   \]
   \[
   12^{12} = (2^2 \cdot 3)^12 = 2^{24} \cdot 3^{12}
   \]

3. Let \(k\) be expressed in terms of its prime factors as \(k = 2^a \cdot 3^b\), where \(a\) and \(b\) are non-negative integers.

4. The LCM of \(6^6\), \(8^8\), and \(k\) is given by:
   \[
   \text{LCM}(6^6, 8^8, k) = \text{LCM}(2^6 \cdot 3^6, 2^{24}, 2^a \cdot 3^b)
   \]

5. The LCM of several numbers is found by taking the highest power of each prime that appears in any of the factorizations. Therefore:
   \[
   \text{LCM}(2^6 \cdot 3^6, 2^{24}, 2^a \cdot 3^b) = 2^{\max(6, 24, a)} \cdot 3^{\max(6, 0, b)}
   \]

6. For the LCM to be equal to \(12^{12} = 2^{24} \cdot 3^{12}\), we must have:
   \[
   2^{\max(6, 24, a)} = 2^{24} \quad \text{and} \quad 3^{\max(6, 0, b)} = 3^{12}
   \]

7. From \(2^{\max(6, 24, a)} = 2^{24}\), it follows that:
   \[
   \max(6, 24, a) = 24 \implies a \leq 24
   \]

8. From \(3^{\max(6, 0, b)} = 3^{12}\), it follows that:
   \[
   \max(6, 0, b) = 12 \implies b = 12
   \]

9. Therefore, \(a\) can take any integer value from \(0\) to \(24\), inclusive. This gives us \(25\) possible values for \(a\).

10. Since \(b\) must be exactly \(12\), there are no other constraints on \(b\).

Thus, there are \(25\) possible values for \(k\).

The final answer is \(\boxed{25}\).