to find the number of trailing 0's in 28! I first found the prime factorization of 28! which comes to $28! = 2^{25} \times 3^{13} \times 5^6 \times 7^4 \times 11^2 \times 13^2 \times 17 \times 19 \times 23$
the prime factors of 12 are $3 \times 2^2$
but to try and get $2^{25} \times 3^{13}$ in terms of $12^n$ I get $(2^{24} \times 3^{12}) \times 2 \times 3$ which equals $12^{12}$ $\times 2 \times 3$
does this mean the number of 0's trailing 28! in base 12 is 12?