multiples

$$12/3 = 4, 12/9 ≈ 1.$$

There are 5 powers of 3 in $12!$

$$15/3=5, 15/9 ≈1.$$

There are 6 powers of 3 in $15!$

$$18/3 = 6, 18/9 =2.$$

There are 8 powers of 3 in $18!$

You can likewise figure out that there are 9 in $21!$ and 10 in $24!.$ Thus, as there are 5 numbers, there will be no multiplying of 3 within non-3-divisible numbers, so the answer will relate to the sum $3^5 + 3^6 + 3^8 + 3^9 + 3^{10}.$ This is just $3^5(1 + 3 + 27 + 81 + 243,)$ or $3^5(355),$ so the answer is $\boxed{5}.$