The question is basically asking us to determine how many 3's we can factor out of \(9! – 6^4\)
This calls for some prime factorization!

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Here's a quick technique for finding the prime factorization of a factorial.

For this example, we'll find the prime factorization of 10!

First, I know that the prime numbers "hiding" in 10! will range from 2 to 7 (i.e., all primes less than or equal to 10), which means we can write: \(10! = (2^?)(3^?)(5^?)(7^?)\)
From here, I'll focus on one prime number at a time, starting with 2.
In 10!, we have:
1 2 hiding in 2
2 2's hiding in 4 (since 4 = 2x2)
1 2 hiding in 6 (since 6 = 2x3)
3 2's hiding in 8 (since 8 = 2x2x2)
1 2 hiding in 10 (since 10 = 2x5)
So, there's a total of 8 2's hiding in 10!, which means we can now write: \(10! = (2^8)(3^?)(5^?)(7^?)\)

Now let's focus on how many 3's are hiding in 10!.
We have:
1 3 hiding in 3
1 3 hiding in 6 (since 6 = 2x3)
2 3's hiding in 9 (since 9 = 3x3)
So, there's a total of 4 3's hiding in 10!, which means we can write: \(10! = (2^8)(3^4)(5^?)(7^?)\)

Onto the 5's:
1 5 hiding in 5
1 5 hiding in 10
So, there's a total of 2 5's hiding in 10!, which means we can write: \(10! = (2^8)(3^4)(5^2)(7^?)\)

Finally, we can see that there's exactly 1 7 hiding in 10!, which means we can write: \(10! = (2^8)(3^4)(5^2)(7^1)\)
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Following the same technique for \(9!\), we get: \(9! = (2^7)(3^4)(5^1)(7^1) = 2^7 \cdot 3^4 \cdot 5 \cdot 7 \)

Now let's find the prime factorization of \(6^4\) to get: \(6^4 = (2 \cdot 3)^4 = 2^4 \cdot 3^4\)

So, we have: \(n = 9! – 6^4\) \(= 2^7 \cdot 3^4 \cdot 5 \cdot 7 - 2^4 \cdot 3^4\)

Factor the right side to get: \(= 2^4 \cdot 3^4(2^3 \cdot 5 \cdot 7 - 1)\)

Evaluate the part in parentheses: \(= 2^4 \cdot 3^4(3^2 \cdot 31)\)

Simplify one last time: \(= 2^4 \cdot 3^6 \cdot 31\)

So, in total, we can factor six 3's out of \(9! – 6^4\)

Answer: D

Cheers,
Brent