parkhydel wrote:
If \(n = 9! – 6^4\), which of the following is the greatest integer k such that \(3^k\) is a factor of n ?
A. 1
B. 3
C. 4
D. 6
E. 8
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!--------------ASIDE---------------------------------------------------------------------------------------------------------------------------------
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: \(10! = (2^8)(3^4)(5^2)(7^1)\)
-------BACK TO THE QUESTION---------------------------------------------------------------------------------------------------------------------
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:
\(= 2^4 \cdot 3^4(279)\)Find the prime factorization of 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
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Brent Hanneson – Creator of gmatprepnow.com
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