If n = 9! – 6^4, which of the following is the greatest integer k such

Question

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

PS66661.02

BrentGMATPrepNow · Accepted Answer

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

lacktutor · Answer

n= 9! —6^{4} = 2^{7}*3^{4}*5*7 —2^{4}*3^{4} =

=  2^4*3^4 ( 8*5*7–1)

=  2^4*3^{4} *279

279= 3^{2} *31 
—>  2^4* 3^4* 3^2* 31

3^k = 3^6  
—>  k= 6

Answer (D)

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nick1816 · Answer

Maximum number of 2's in 9! =   + +  = 7

Maximum number of 3's in 9! =   +  = 4

We can write 9! as  2^7*3^4*5*7

n =   -

n = 2^4*3^4

n= 2^4*3^4

n = 2^4*3^6*31

D

MentoroniStudents · Answer

The idea is to see how many '3's are there is the equation: 9! – 6^4

9*8*7*6*5*4*3*2 - 3*2*3*2*3*2*3*2 = 3*3*2*2*2*7*3*2*5*2*2*3*2 = 3^4 * 2^4 (7*5*2*2*2 - 1) = 3^4 * 2^4 (280-1) = 3^4 * 2^4 (279) = 3^4 * 2^4 (3*3*31) = 3^6 * 2^4 * 31

6 is the answer (Answer: D)

Give Kudos if you like the explanation.

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TarunKumar1234 · Answer

If n=9!–6^4, which of the following is the greatest integer k such that 3^k is a factor of n ?

Now, n=9!–6^4 = 9*8*7*..... - 3^4*2^4 = 3^4 (8*7*2*5*4*2*1 - 2^4) = 3^4* 2^4 (7*5*8-1) = 3^4* 2^4* 279 = 3^4* 2^4*(3*3*31)

So, I think D.

w4rlk · Answer

n=9!-(6^4)

9!=9x8x7x6x5x4x3x2
9!=6x6x6x6x8x7x5
9!=(6^4)x280

n=  - (6^4)
n=(6^4)(280-1)
n=(2^4)(3^4)(279)
n=(2^4)(3^4)(3^2)(13)
n=(2^4)(3^6)(13)

Therefore greatest integer k=6

errorlogger · Answer

https://i.imgur.com/Prwdr7N.jpg

chetan2u ,  IanStewart :
Please help me understand what is wrong with my approach?

BrentGMATPrepNow · Answer

One way to see the error is to try solving a similar question using this technique used...

If n = 20 - 11, what is the greatest integer k such that  3^k  is a factor of n? 
There are zero 3's in the factorization of 20, and there are zero 3's in the factorization of 11. 
So, applying your technique, the number of 3's in the factorization of 20 - 11 equals 0 - 0, which is not the case

errorlogger · Answer

Hey  BrentGMATPrepNow ,
thanks for the quick revert.
However, I still don't get what is wrong with the approach. As per the little knowledge that I have, I believe the number of 3's can be found out using the method that I have shown. It only applies to factorials not normal numbers.
For ex: 3! will have one 3
9! will have 9/3 = 3 + 9/9 = 1, four 3's.
I know I'm wrong somewhere but still struggling to find it. Thanks again!

After re-reading your explanation, I guess you are referring to the split up that I have done and that is the root cause?

BrentGMATPrepNow · Answer

The approach doesn't work with factorials either. 
The original question doesn't ask us to divide by 3^k. You inserted that in order to apply a technique for determining the number of 3's "hiding" in a number.

Here's an example to show that the approach doesn't work with factorials: 
 If n = 4! - 2!, what is the greatest integer k such that  2^k  is a factor of n? 
There are 3 twos in the factorization of 4!, and there is 1 two in the factorization of 2!. 
So, applying your technique, the number of 2's in 4! - 2! equals 3 - 1 = 2. 
However, 4! - 2! = 22, and there's only 1 two in the factorization of 22

IanStewart · Answer

Since 9! is divisible by 3^4, and 6^4 is divisible by 3^4, then you can be completely certain you'll be able to factor 3^4 out of 9! + 6^4, or from 9! - 6^4, or even from 9! + (11)(6^4), and so on. But all that proves is that we can divide 9! - 6^4 by  at least  3^4. We might still be able to divide it by 3^5 or 3^6 or a higher power of 3, because new factors can show up when you add or subtract. For example, 27 and 54 are both divisible by 3^3, but when you add them, you get 81, which is divisible by 3^4.

So you've completed the first step of the problem, but then you genuinely need to factor out the 3^4 and see how many 3's you get in the brackets. Then you'll get the right answer.

chetan2u · Answer

Hi,

You already have the relevant answers by experts above.

Whenever you are looking at remainders or other properties, tackle the entire expression rather than individual terms in it. 
Adding the term may lead to an extra 3 or lesser number of 3s too.

Example:
3^4+6 will have only 3^1 as its factor. 
While 3^3+6^3 may have 3^3 as factors of each terms - 3^3 and 6^3, when combined they give you 2 extra 3s, that is even 3^5 is a factor.

gauravjain0211 · Answer

What leads to error in such a problem - your confidence cause you already know the concept involved in calculation of the power of 3 in a factorial (too easy to solve right, except it isn't)
 
so yes 9! is div by at max 3^4, but that subtraction must induce a suspicion and persuade you to double check (cause even when individual numbers don't have 3 in them as a factor, their addition or subtraction may lead to a multiple of 3 for eg - (11-2)

So now factorial is not too big so open and do the exact calculation

1*2*3*4*5*6*7*8*9 - 6^4

6^4 (1*5*7*8 - 1) (cause 2*3*4*9 = 6^3)

6^4 (280-1)
6^4 (279)
6^4 (31*9)
6^4(31*3^2)

hence 3^6 can divide it, hence k = 6

KarishmaB · Answer

Keep in mind that addition or subtraction of terms can bring in new factors which don't exist in the two terms.
For example, if a and b are co-prime such as 3 and 5, 6a ( = 18) and 6b (= 30) have only one 2 each.

But  6a + 6b = 6(a+b) = 6 * (3 + 5) = 6 * 8  has four 2s.

Similarly,  (3^4 * 10 - 3^4)  has only four 3s in each term but their subtraction leads to two more 3s.
 3^4 (10 - 1) = 3^4 * 9

Hence, the only trick in these questions is to calculate what is obtained by the addition/subtraction and find its factors.

jaydavido · Answer

hi, I know this will sound stupid, put what do you mean with factoring the right side, I really dont know how to do that, can u explain?

Anonymous · Answer

Could you clarify how you derived 9! as 6x6x6x6x8x7x5 from 9x8x7x6x5x4x3x2? Also, in the highlighted text, should it not be 3^2 * 31 since 31 multiplied by 9 equals 279?

I also arrived at k = 6, but my method was as follows: I identified the multiples of 3 (since we're looking for 3^k) within 9! and divided each by 3 to determine the number of 3s present in them.

9 divided by 3 equals 3,
6 divided by 3 equals 2,
3 divided by 3 equals 1.

Summing 3+2+1 yields 6, which equals k.

Is this method correct?

nekr0 · Answer

Prime factorise 9! and it'll generate  (3^4)(2^7)(7*5)  =  (3^4 * 2^4) * (2^3) * (7 * 5)  =  (3 * 2)^4 * (8 * 7 * 5)  =  6^4 * 280 .

As for the last part - yes, it should've been  (3^2 * 31)  and '13' is a typo, but it wouldn't affect the results as both are primes.

n =  9! – 6^4 
n =    - (6^4) 
n =  (6^4)(280-1) 
n =  (2^4)(3^4)(279) 
n =  (2^4)(3^4)(3^2 * 31) 
n =  2^4 *(3^4)(3^2) *31  
n =  (3^6) * 2^4 * 31

Thus, k(max) = 6

thekingslay · Answer

Solution-

9!-6^4/3
or, rewriting - 9/3 - 6^4/3 - quotients are 3 and 2
then, 9/3^2 - 6*6*6*6/3^2 - quotients are 1 and 1
then, 9/3^3 can not be solved, so focusing on 6*6*6*6/3^3 - quotient is 1
then, 6*6*6*6/3^4 - quotient is 1. 
so the final power to solve this is 2 (from the first part)+4(from the other part) = 6

sagar123450909 · Answer

n = 9! - 6^4

= 9*8*7*6*5*4*3*2 - 6^4

= 3*3*2*2*2*7*6*5*4*3*2 - 6^4

= 6^4 (2*7*5*4 - 1)

= 6^4 (280-1)

= 6^4 (279)

= 6^4*3*3(31)

= 3^6 * 2^4 * 31

Hence greatest integer k such that n is divisible 3^k is 6

If n = 9! – 6^4, which of the following is the greatest integer k such

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GMAT Problem Solving (PS) Questions

If n = 9! – 6^4, which of the following is the greatest integer k such

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