In the prime factorization of 12! + 15! + 18! + 21! + 24!, what is the exponent of 3?
]12! + 15! + 18! + 21! + 24!]= 2^10 * 3^5 * 5^2 * 7 * 11 * 53 * 129169 * 189221423
12! + 15! + 18! + 21! + 24! = 12!(1 + 15...13 + 18...13 + 21...13 + 24...13)
Each of the four products within the brackets is a multiple of 3, so their sum will be a multiple of 3, meaning that the sum of the five numbers within the brackets will be one more than a multiple of 3 and so will not contain 3 as a prime factor.
Any 3 has to come from the 12!
So, 3 -> 1, 6 -> 1, 9 -> 2, 12 -> 1,
five in all.
