In the prime factorization of 12! + 15! + 18! + 21! + 24!, what is the exponent of 3?
+421 $$12/3 = 4, 12/9 ≈ 1.$$
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 $\boxed{5}.$
