jmc

It is evident that $k$ has only 2s and 3s in its prime factorization, or $k=2^a{3}^b$. $6^6=2^6\cdot 3^6$ $8^8=2^{24}$ $12^{12}=2^{24}\cdot 3^{12}$ The LCM of any numbers an be found by writing out their factorizations and taking the greatest power for each factor. $[6^6,8^8]=2^{24}{3}^6$. Therefore $12^{12}=2^{24}\cdot 3^{12}=[2^{24}{3}^6,2^a{3}^b]=2^{\max (24,a)}{3}^{\max (6,b)}$, and $b=12$. Since $0\le a\le 24$, there are $\boxed{25}$ values of $k$.