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